RS Aggarwal 2020 2021 Solutions for Class 7 Maths Chapter 17 - Constructions

Question 1:

Question 2:

$$\begin{gathered} \begin{array}{l}\text{Steps of construction:}\\ 1.\mathrm{Draw} \mathrm{a} \mathrm{line} \mathrm{AB}.\end{array} \\ \begin{array}{l}2. \text{Take a point Q on AB and a point P outside AB, and join PQ.}\\ \text{3. With Q as the centre and any radius, draw on arc to cut AB at X and PQ at Z.}\\ \text{4. With P as the centre and the same radius, draw an arc cutting QP at Y .}\\ \text{5. With Y as the centre and the radius equal to XZ, draw an arc to cut the previous arc at E.}\\ \text{6. Join PE and produce it on both the sides to get the required line}\text{.}\end{array} \end{gathered}$$

Question 3:

$\begin{array}{l}\text{Steps for construction:}\\ 1. \text{Let AB be the given line}\text{.}\\ 2. \text{Take any two points P and Q on AB}\text{.}\\ \text{3}\text{. Construct }\angle BPE=90°\text{ and }\angle BQF=90°\\ 4. \text{With P as the centre and the radius equal to 3}\text{.5 cm, cut PE at R.}\\ \text{5}\text{. With Q as the centre and the radius equal to 3}\text{.5cm, cut QF at S.}\\ \text{6}\text{. Join RS and produce it on both the sides to get the required line, parallel to AB and at a distance of 3}\text{.5 cm from it}\text{.}\\ \\ \end{array}$

Question 1:

$\begin{array}{l}\text{Steps of construction:}\\ \text{1}\text{. Let}\mathit{\text{ l}}\text{ be the given line}\text{.}\\ \text{2}\text{. Take any two points A and B on line}\mathit{\text{ l.}}\\ 3. \text{Construct }\angle B\text{AE}=90°\text{ and }\angle \text{ABF=90}°\\ 4. \text{With A as the centre and the radius equal to 4}\text{.3 cm, cut AE at C}\text{.}\\ \text{5}\text{. With B as the centre and the radius equal to 4}\text{.3 cm, cut BF at D}\text{.}\\ \text{6}\text{. Join CD and produce it on either side to get the required line m, parallel to l and at a distance of 4}\text{.3 cm from it}\text{.}\\ \\ \end{array}$

Question 2:

$$\begin{gathered} \begin{array}{l}\text{Steps of construction:}\\ \text{ 1}\text{. Draw a line segment (AB) of length 5 cm.}\\ \text{ 2}\text{. Draw an arc of radius 5}\text{.4 cm from the centre (A).}\\ \text{ 3}\text{. With B as the centre, draw another arc }\mathrm{of radius 3}\text{.6 cm, cutting the previous arc at C. }\\ \text{ 4}\text{. Join AC and BC}\text{.}\\ \text{ 5}\text{. Taking B as the centre and the radius more than half of BC, draw two arcs on both the sides of BC}\text{.}\\ \text{ 6}\text{. Similarly, taking C }\text{}\text{as the centre and the same radius, draw arcs on both the sides of BC, cutting the previous arcs at P and Q.}\\ \text{ 7. Join PQ. }\end{array} \\ \text{Then, PQ is the required perpendicuar bisector of BC, meeting BC at D.} \end{gathered}$$

Question 3:

$$\begin{gathered} \begin{array}{l}\text{Steps of construction:}\\ 1. \text{Draw a line segment QR of length 6 cm}\text{.}\\ \text{2}\text{. Draw arcs of 4}\text{.4 cm and 5}\text{.3 cm from Q and R, respectively. They intersect at P.}\\ \text{3}\text{. Draw an arc of any radius from the centre (P), cutting PQ and PR at S and T, respectively.}\\ \text{4}\text{. With S as the centre and the radius more than half of ST, draw an arc}\text{ .}\\ \text{5. With T as the centre and the }\mathrm{same radius}, draw \mathrm{another} \mathrm{arc} \mathrm{cutting} \mathrm{the} \mathrm{previously} \mathrm{drawn} \mathrm{arc} \mathrm{at} \mathrm{X}.\\ \text{6. Join P and X}\text{.}\end{array} \\ \mathrm{Then}, \mathrm{PX} \mathrm{is} \mathrm{the} \mathrm{bisector} \mathrm{of} \angle \mathrm{P}. \end{gathered}$$

Question 4:

$\begin{array}{l}\text{Steps of construction:}\\ 1. \text{Draw AB of length 6}\text{.2 cm}\text{.}\\ \text{2}\text{. By taking the centres as A and B, draw equal arcs of 6}\text{.2 cm on the same side of AB, cutting each other at C}\text{.}\\ \text{3}\text{. Join AC and BC}\text{.}\\ \end{array}$
 
When we will measure angles of triangle using protractor then we find that all angles are equal to 60$°$

Question 5:

$\begin{array}{l}\text{Steps of construction:}\\ \text{1}\text{. Draw BC=5}\text{.3 cm}\\ \text{2}\text{. Draw an arc of radius 4}\text{.8 cm from the centre, B}\text{.}\\ 3. \text{Draw another arc of radius 4}\text{.8 cm from the centre, C}\text{.}\\ \text{4}\text{. Both of these arcs intersect at A}\text{.}\\ \text{5}\text{. Join AB and AC}\text{.}\\ \text{6}\text{. With A as the centre and any radius, draw an arc cutting BC at M and N. }\\ \\ \text{7}\text{. With M as the centre and the radius more than half of MN, draw an arc.}\\ \text{8. With N as the centre and the same radius, draw another arc cutting the previously drawn arc at P. }\\ \text{9. Join AP, cutting BC at D.}\\ \mathrm{Then}, AD\perp BC\end{array}$

Question 6:

$$\begin{gathered} \begin{array}{l}\text{Steps of construction:}\\ \text{1}\text{. Draw AB of length 3}\text{.8 cm}\text{.}\\ \text{2}\text{. Draw ∠BAZ=60°}\\ \text{3}\text{. With the centre as A, cut ray AZ at 5 cm at C.}\\ \text{4 Join BC.}\end{array} \\ \text{Then, ABC is the required triangle.} \end{gathered}$$

Question 7:

$\begin{array}{l}\begin{array}{l}\begin{array}{l}\text{Steps of construction:}\\ \text{1. Draw AC= 6 cm}\end{array}\\ \text{2. }\mathrm{Draw} \angle \mathrm{ACZ}=45°\end{array}\\ 3. \mathrm{With} \mathrm{C} \mathrm{as} \mathrm{the} \mathrm{centre}, \mathrm{cut} \mathrm{ray} \mathrm{BZ} \mathrm{at} 4.3 \mathrm{cm} \mathrm{at} \mathrm{point} \mathrm{B}.\\ \begin{array}{l}\text{4. Join AB.}\\ \mathrm{Then}, \mathrm{ABC} \mathrm{is} \mathrm{the} \mathrm{required} \mathrm{triangle}.\end{array}\end{array}$

Question 8:

$$\begin{gathered} \begin{array}{l}\text{Steps of construction:}\\ \text{1}\text{. Draw AB=5.2 cm}\\ \text{2}\text{. Draw ∠BAX=120° }\\ \text{3}\text{. With A as the centre, cut the ray AX at 5.3 cm at point C.}\\ \text{4}\text{. Join BC.}\\ \text{5}\text{. With A as the centre and any radius, draw an arc cutting BC at M and N}.\\ 6. \text{With M as the centre and the radius more than half of MN, draw an arc.}\\ \text{7}\text{. With N as the centre and the same radius as before, draw another arc cutting the previously drawn arc at P.}\\ 8. \mathrm{Join} \mathrm{AP} \mathrm{meeting} \mathrm{BC} \mathrm{at} \mathrm{D}.\end{array} \\ \therefore \mathrm{AD}\perp \mathrm{BC} \end{gathered}$$

Question 9:

$$\begin{gathered} \begin{array}{l}\text{Steps of construction:}\\ \text{1}\text{. Draw BC=6.2 cm}\\ \text{2}\text{. Draw ∠BCX=45°}\\ \text{3}\text{. Draw ∠CBY=60°}\end{array} \\ 4.\mathrm{The} \mathrm{ray} \mathrm{CX} \mathrm{and} \mathrm{BY} \mathrm{intersect} \mathrm{at} \mathrm{A}. \\ \mathrm{Then}, \mathrm{ABC} \mathrm{is} \mathrm{the} \mathrm{required} \mathrm{triangle}. \end{gathered}$$

Question 10:

$$\begin{gathered} \begin{array}{l}\text{Steps of construction:}\\ \text{1}\text{. Draw BC=5}\text{.8 cm}\end{array} \\ 2. \mathrm{Draw} \angle \mathrm{BCY}=30° \\ 3. \mathrm{Draw} \angle \mathrm{CBX}=30° \\ 4. \mathrm{The} \mathrm{ray} \mathrm{BX} \mathrm{and} \mathrm{CY} \mathrm{intersect} \mathrm{at} \mathrm{A}. \\ \mathrm{Then}, \mathrm{ABC} \mathrm{is} \mathrm{the} \mathrm{required} \mathrm{triangle}. \\ \mathrm{On} \mathrm{measuring} \mathrm{AB} \mathrm{and} \mathrm{AC}: \\ \mathrm{AB}=\mathrm{AC}=3.4 \mathrm{cm} \end{gathered}$$

Question 11:

$$\begin{gathered} \mathrm{By} \mathrm{angle} \mathrm{sum} \mathrm{property}: \\ \angle B=180°-\angle A-\angle C \\ =180°-45°-75° \\ =60° \\ \begin{array}{l}\text{Steps of construction:}\\ \text{1}\text{. Draw AB=7cm}\\ \text{2 Draw ∠BAX= 45°}\\ 3. \text{Draw ∠ABY= 60°}\\ \text{4.The ray AX and BY intersect at C.}\end{array} \\ \text{Then, ABC is the required triangle.} \end{gathered}$$

Question 12:

$\begin{array}{l}\text{Steps of construction:}\\ \text{1}\text{.Draw BC=4}\text{.8 cm}\\ \text{2}\text{.Draw a perpendicular on C such that }\angle {\text{C is equal to 90°.}}^{}\\ 3.\text{Draw an arc of radius 6}\text{.3 cm from the centre B.}\\ 4.\text{Join AB.}\end{array}$

Question 13:

$$\begin{gathered} \begin{array}{l}\text{Steps of construction:}\\ \text{1}\text{. Draw AB=3}\text{.5 cm}\\ \text{2}\text{. Construct }\angle \text{ABX}=90°\\ 3. \mathrm{With} \mathrm{centre} \mathrm{A}, \mathrm{d}\text{raw an arc of radius 6 cm cutting BX at C.}\\ \text{4}\text{. Join AC.}\end{array} \\ \text{Then, ABC is the required triangle.} \end{gathered}$$

Question 1:

$$\begin{gathered} \begin{array}{l}\text{Here, }\angle \text{A=30° and }\angle \text{C=90}°\\ \text{By angle sum property: }\\ \angle \text{B=60}°\end{array} \\ {\text{ 1. Draw the hypotenuse AB of length 5.6 cm.}}^{} \\ \text{2. Draw}\angle BAX=30° \mathrm{and} \angle ABY=60° \\ 3. \mathrm{The} \mathrm{ray} \mathrm{AX} \mathrm{and} \mathrm{BY} \mathrm{intersect} \mathrm{at} \mathrm{C}. \\ \mathrm{Then}, \mathrm{ABC} \mathrm{is} \mathrm{the} \mathrm{required} \mathrm{triangle}. \end{gathered}$$

Question 2:

$$\begin{gathered} \left(\mathrm{c}\right) 135° \\ \begin{array}{l}\\ {\text{ Supplement of 45}}^{°}{\text{=180}}^{°}-{45}^{°}\\ {\text{ =135}}^{°}\\ \end{array} \end{gathered}$$

Question 3:

$$\begin{gathered} \left(\mathrm{b}\right) 10° \\ \begin{array}{l}{\text{Complement of 80}}^{°}\text{ = }{90}^{°}-{80}^{°}\\ {=10}^{°}\end{array} \end{gathered}$$

Question 4:

$$\begin{gathered} \begin{array}{l}\left(\text{b}\right)45°\end{array} \\ \mathrm{Suppose} \mathrm{the} \mathrm{angle} \mathrm{is} x°. \\ \mathrm{Then}, \mathrm{the} \mathrm{complement} \mathrm{is} \mathrm{also} x°. \\ \mathrm{Complement} \mathrm{of} x°=90°-x° \\ \Rightarrow x°=90°-x° \\ \Rightarrow x°+x°=90° \\ \Rightarrow 2x°=90° \\ \Rightarrow x=\frac{90}{2} \\ \Rightarrow x=45 \end{gathered}$$

Question 5:

$$\begin{gathered} \left(a\right) 30° \\ \mathrm{Suppose} \mathrm{the} \mathrm{angle} \mathrm{is} x. \\ x=\frac{(180-x)}{5} \\ \Rightarrow 5x=180-x \\ \Rightarrow 5x+x=180 \\ \Rightarrow x=\frac{180}{6} \\ \Rightarrow x=30{°}^{} \end{gathered}$$

Question 6:

$$\begin{gathered} \left(b\right) 57° \\ S\mathrm{uppose} \mathrm{the} \mathrm{angle} \mathrm{is} \mathrm{x}. \\ x=90-x+24 \\ \Rightarrow x+x=114 \\ \Rightarrow 2x=114 \\ \Rightarrow x=\frac{114}{2} \\ \Rightarrow x=57{°}^{} \end{gathered}$$

Question 7:

$$\begin{gathered} \left(b\right) 74° \\ \mathrm{Suppose} \mathrm{the} \mathrm{angle} \mathrm{is} x. \\ x=180-x-32 \\ \Rightarrow x+x=148 \\ \Rightarrow 2x=148 \\ \Rightarrow x=\frac{148}{2} \\ \Rightarrow x=74° \end{gathered}$$

Question 8:

$$\begin{gathered} \left(c\right) 72° \\ \begin{array}{l}\text{Supplementary angles:}\\ 3x+2x=180\\ \text{ =>}x=\frac{180}{5}\\ \end{array} \\ \Rightarrow x=36° \\ \begin{array}{l}\text{ Smaller angle = }(2\times 36°)\\ \text{ =72}{°}^{}\end{array} \end{gathered}$$

Question 9:

$$\begin{gathered} \left(b\right) 48° \\ \angle AOC+\angle BOC=180° (\mathrm{linear} \mathrm{pair}) \\ \angle AOC=180°-\angle BOC \\ =180°-132° \\ =48° \end{gathered}$$

Question 10:

$$\begin{gathered} \left(\mathrm{x}\right) 112 \\ \angle AOC+\angle AOB =180° (\mathrm{linear} \mathrm{pair}) \\ 68°+\mathrm{x}°=180° \\ \Rightarrow x°=180°-68° \\ \Rightarrow x°=112° \end{gathered}$$

Question 11:

$\begin{array}{l}\left(\mathrm{c}\right)x=35\\ \\ (2x-10)+(3x+15)=180\\ =>2x-10+3x+15=180\\ =>5x+5=180\\ =>5x=180-5\\ =>5x=175\\ =>x=\frac{\sout{1}\sout{7}{\sout{5}}^{35}}{{\sout{5}}^1}\\ =>x=35\end{array}$

Question 12:

$$\begin{gathered} \left(d\right) x=80 \\ x+55+45=180 (\mathrm{linear} \mathrm{pair}) \\ \Rightarrow \mathrm{x}=180-55-45 \\ \Rightarrow \mathrm{x}=180-100 \\ \Rightarrow \mathrm{x}=80 \end{gathered}$$

Question 13:

$$\begin{gathered} \left(a\right) 100 \\ \begin{array}{l}x+y=180 (\mathrm{linear} \mathrm{pair})\\ \text{ =>}x+\frac{4}{5}x=180°\\ \text{ =>9}x=5\times 180\\ \text{ =>}x=100\\ \end{array} \end{gathered}$$

Question 14:

$$\begin{gathered} \left(\mathrm{b}\right) 50° \\ \begin{array}{l}\text{ Here, }\angle \text{AOC and }\angle \text{BOD are vertically opposite angles}\text{.}\\ \therefore \angle \text{AOC=}\angle \text{BOD}\\ \text{ Given, }\angle {\text{AOC=50}}^0\\ \therefore \angle {\text{BOD=50}}^0\\ \end{array} \end{gathered}$$

Question 15:

$$\begin{gathered} \left(a\right) 32 \\ \begin{array}{l}\text{ (3}x-8)°+(x+10)°+50°=180° (\mathrm{linear} \mathrm{pair})\\ \text{ =>4}x°+52°=180°\\ \text{ =>4}x°=128°\\ \text{ =>}x°=32°\\ \end{array} \\ \therefore x=32 \end{gathered}$$

Question 16:

$$\begin{gathered} \left(\mathrm{b}\right) 78° \\ \begin{array}{l}\angle \text{ACD=}\angle \text{ABC+}\angle \text{BAC (exterior angle property)}\\ \text{ =>}\angle \text{ABC=132°}-54°=78°\end{array} \end{gathered}$$

Question 17:

$$\begin{gathered} \left(\mathrm{c}\right) 100° \\ \begin{array}{l}\angle \text{ACB}=\angle \text{ABC+}\angle \text{BAC (exterior angle property)}\\ \text{ =}(45°+55°)\\ \text{ =100}°\\ \\ \end{array} \end{gathered}$$

Question 18:

$$\begin{gathered} \left(\mathrm{b}\right) 50° \\ \begin{array}{l}\angle \text{BCA}={180}^0-{120}^0 (\mathrm{linear} \mathrm{pair})\\ ={60}^0\\ \angle \text{BAC}={180}^0-({60}^0+{70}^0) (\mathrm{angle} \mathrm{sum} \mathrm{property} \mathrm{of} \mathrm{triangles})\\ ={50}^0\\ \end{array} \end{gathered}$$

Question 19:

$$\begin{gathered} \left(c\right) 150° \\ \begin{array}{l}{x}^0+{70}^0+{50}^0+{90}^0={360}^0 (\mathrm{complete} \mathrm{angle})\\ =>x^0={360}^0-{210}^0\\ ={\text{ 150}}^0\\ \end{array} \end{gathered}$$

Question 20:

$$\begin{gathered} \left(c\right)70° \\ \begin{array}{l}\text{ Here, }\angle \text{ACE}=\angle \text{BAC}={50}^0\left[\text{ alternate angles}\right]\\ \text{ ∠ACB+∠ACE+∠DCE=180° (linear pair)}\\ \angle \text{ACB}={180}^0-(50°+60°)\\ =180°-110°\end{array} \\ =70° \end{gathered}$$

Question 21:

$$\begin{gathered} \left(b\right) 30° \\ \begin{array}{r}\angle \text{A+}\angle \text{B+}\angle \text{C}={180}^0\\ =>\angle \text{B}={180}^0-({65}^0+{85}^0)\\ =>\angle \text{B}={180}^0-{150}^0\\ =>\angle \text{B}={30}^0\\ \\ \end{array} \end{gathered}$$

Question 22:

$\left(\text{d}\right){\text{ 180}}^0$

Question 23:

$\begin{array}{l}{\text{(c) 360}}^0\\ \end{array}$

Question 24:

$$\begin{gathered} \left(b\right) 90° \\ \begin{array}{l}\text{ Draw a parallel line through O and produce AB and CD on R and P, respectively}\text{.}\\ \\ \therefore \angle \text{OCD}=\angle {\text{COQ=120}}^0\text{ (alternate angles)}\end{array} \\ \angle {\text{COS=180}}^0-{120}^0 (\mathrm{linear} \mathrm{pair}) \\ ={60}^0 \\ \text{Similarly, }\angle \text{AOQ=}\angle {\text{BAO=150}}^0 (alternate angles) \\ \angle {\text{AOS=180}}^o-{150}^0 (\mathrm{linear} \mathrm{pair}) \\ ={30}^0 \\ \angle AOC=\angle AOS+\angle COS \\ \therefore \angle {\text{AOC=60}}^0+{30}^0={90}^0 \end{gathered}$$

Question 25:

$$\begin{gathered} \left(\mathrm{a}\right) 40° \\ \begin{array}{l}\angle \text{PAC}=\angle \text{ACS}={100}^0\left[\text{alternate angles}\right]\\ \angle \text{PAB}+\angle \text{BAC}={100}^0\\ =>\angle \text{BAC}=100°-60°=40°\end{array} \end{gathered}$$

Question 26:

$$\begin{gathered} \left(c\right) 30 \\ \begin{array}{l}\text{ Here, }\angle \text{DCG}+\angle \text{CGF}={180}^0 (\mathrm{angles} \mathrm{on} \mathrm{the} \mathrm{same} \mathrm{side} \mathrm{of} \mathrm{a} \mathrm{transversal} \mathrm{line} \mathrm{are} \mathrm{supplementary})\\ =>\angle \text{CGF}={180}^0-100°=80°\\ \angle \text{ABG}=\angle \text{BGF}={110}^0\left[\text{alternate angles}\right]\\ x^0+\angle \text{CGF}={110}^0\\ =>x^0={110}^0-{80}^0\\ =>x^0={30}^0\end{array} \\ \therefore x=30 \end{gathered}$$

Question 27:

$\left(\mathrm{d}\right) \mathrm{greater} \mathrm{than} \mathrm{the} 3\mathrm{rd} \mathrm{side}$

Question 28:

$\left(\text{d}\right)\text{ The diagonals of a rhombus always bisect each other at right angles}\text{.}$

Question 29:

$$\begin{gathered} \left(c\right) 12 \mathrm{cm} \\ \begin{array}{l}\text{ In a right angle triangle: }\\ {\text{ AC}}^2={\text{AB}}^2+{\text{BC}}^2\text{ (Pythagoras theorem)}\\ =>{\text{BC}}^2={13}^2-5^2\\ {\text{ => BC}}^2=169-25\\ =>{\text{BC}}^2=144\\ =>\text{BC =±12 }\end{array} \\ \text{The length cannot be negative.} \\ \text{ ∴ BC= 12 cm} \end{gathered}$$

Question 30:

$$\begin{gathered} \left(c\right) 114° \\ \begin{array}{l}\text{ In triangle ABC:}\\ \angle \text{A+}\angle \text{B+}\angle {\text{C=180}}^0\\ =>\angle {\text{A=180}}^0-({37}^0+{29}^0)\\ =>\angle {\text{A=180}}^0-\left({66}^0\right)\\ ={114}^0\\ \end{array} \end{gathered}$$

Question 31:

$$\begin{gathered} \left(c\right) 105° \\ \mathrm{Suppose} \mathrm{the} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{triangle} \mathrm{are} 2x, 3x \mathrm{and} 7x. \\ \begin{array}{l}\text{ Sum of the angles of a triangle is 180°.}\\ \text{ 2}x+3x+7x=180\\ =>12x=180\\ =>x={15}^0\\ \text{ Measure of the largest angle = }{15}^0\times 7={105}^0\\ \\ \end{array} \end{gathered}$$

Question 32:

$$\begin{gathered} \left(c\right) 60° \\ Given: \\ 2\angle \text{A}=3\angle B \mathrm{or} \angle \text{A=}\frac{3}{2}\angle B \\ 3\angle \text{B}=6\angle \text{C, or ∠C=}\frac{1}{2}\text{∠B} \\ \begin{array}{l}\text{ In a }△\text{ABC: } \\ \angle \text{A+}\angle \text{B+}\angle \text{C}={180}^0\\ \text{ =>}\frac{3}{2}\angle \text{B+}\angle \text{B +}\frac{1}{2}\angle \text{B}={180}^0\\ \text{ =>}\frac{3\angle \text{B+2}\angle \text{B+}\angle \text{B}}{2}={180}^{{}^0}\\ \text{ =>}\frac{6\angle \text{B}}{2}={180}^0\\ \text{ =>}\angle \text{B=}\frac{{360}^0}{6}\\ \text{ =>}\angle {\text{B=60}}^0\\ \end{array} \end{gathered}$$

Question 33:

$$\begin{gathered} \left(a\right) 25° \\ \mathrm{Given}: \\ \angle A+\angle B=65° \\ \angle A=65°-\angle B ...\left(i\right) \\ \angle B+\angle C=140° \\ \angle C=140°-\angle B ...\left(ii\right) \\ \mathrm{In} ∆\mathrm{ABC}: \\ \angle A+\angle B+\angle C=180° \\ \mathrm{Putting} \mathrm{the} \mathrm{value} \mathrm{of} \angle \mathrm{B} \mathrm{and} \angle \mathrm{C}: \\ \Rightarrow 65°-\angle B+\angle B+140°-\angle B=180° \\ \Rightarrow -\angle B=180°-205° \\ \Rightarrow \angle B=25° \end{gathered}$$

Question 34:

$$\begin{gathered} \left(b\right) 55° \\ \begin{array}{l}\text{ In }△\text{ABC:}\\ \angle \text{A}+\angle \text{B}+\angle \text{C}={180}^0 ...\left(i\right)\\ \text{ Given:}\\ \angle \text{A}-\angle \text{B}={33}^0=>\angle \text{A}=\angle \text{B}+{33}^0 ...\left(ii\right)\\ \angle \text{B}-\angle \text{C}={18}^0=>\angle \text{C}=\angle \text{B}-{18}^0 ...\left(iii\right)\\ \text{Using }\left(ii\right)\text{ and }\left(iii\right)\text{ in equation }\left(i\right):\\ =>\angle {\text{B + 33}}^0+\angle \text{B + }\angle \text{B}-{18}^0={180}^0\\ \text{ => 3}\angle {\text{B + 15}}^0={180}^0\\ \text{ => 3}\angle {\text{B = 165}}^0\\ \text{ => }\angle \text{B = }\frac{{165}^0}{3}={55}^0\\ \\ \end{array} \end{gathered}$$

Question 35:

$$\begin{gathered} \left(\mathrm{c}\right) 22 \\ \begin{array}{l}\text{ Sum of the angles of a triangle is 180°.}\\ (3\mathrm{x})°+(2\mathrm{x}-7)°+(4\mathrm{x}-11)°=180°\\ =>9\mathrm{x}°-18°=180°\\ =>9\mathrm{x}°=198°\\ =>\mathrm{x}°=22°\end{array} \\ \Rightarrow \mathrm{x}=22 \end{gathered}$$

Question 36:

$$\begin{gathered} \left(\mathrm{c}\right) 25 \mathrm{cm} \\ \begin{array}{l}\text{ In a right angle triangle ABC:}\\ {\text{ AC}}^2={\text{BC}}^2+{\text{AB}}^2\\ {\text{ =>BC}}^2={24}^2+7^2\\ {\text{ =>BC}}^2\text{= }576+49\\ {\text{ =>BC}}^2=625\\ \text{ =>BC}=\pm 25\text{ cm}\end{array} \\ \mathrm{Since} \mathrm{the} \mathrm{length} \mathrm{cannot} \mathrm{be} \mathrm{negative}, \mathrm{we} \mathrm{will} \mathrm{negelect} -25. \\ \therefore \mathrm{BC}=25 \mathrm{cm} \end{gathered}$$

Question 37:

$$\begin{gathered} \left(b\right) 25 \mathrm{m} \\ \begin{array}{l}\text{In right triangle ABC:}\\ {\text{ AC}}^2={\text{AB}}^2+{\text{BC}}^2\\ {\text{ =15}}^2+{20}^2\\ {\text{ => AC}}^2=625\\ \text{ =>AC}=\pm 25\\ \mathrm{Since} \mathrm{the} \mathrm{length} \mathrm{cannot} \mathrm{be} \mathrm{negative}, \mathrm{we} \mathrm{will} \mathrm{negelect} -25.\end{array} \\ \therefore \text{ Length of the ladder = 25 m} \end{gathered}$$

Question 38:

$$\begin{gathered} \left(a\right) 13 \mathrm{m} \\ \mathrm{Suppose} \mathrm{there} \mathrm{are} \mathrm{two} \mathrm{poles} \mathrm{AE} \mathrm{and} \mathrm{BD}. \\ \mathrm{EC}=\mathrm{AB}=12 \mathrm{m} (\mathrm{ABCE} \mathrm{is} \mathrm{a} \mathrm{rectangle}) \\ \mathrm{AE}= \mathrm{BC}= 6 \mathrm{m} (\mathrm{ABCE} \mathrm{is} \mathrm{a} \mathrm{rectangle}) \\ \mathrm{DC}= \mathrm{BD}-\mathrm{AE} \\ = 11-6 \\ =5 \mathrm{m} \\ \mathrm{In} \mathrm{the} \mathrm{right} \mathrm{angled} \mathrm{triangle} \mathrm{ECD}: \\ {\mathrm{ED}}^2={\mathrm{EC}}^2+{\mathrm{DC}}^2 (\mathrm{Pythagoras} \mathrm{theorem}) \\ {\mathrm{ED}}^2=5^2+{12}^2 \\ {\mathrm{ED}}^2=25+144 \\ {\mathrm{ED}}^2=169 \\ \mathrm{ED}=\pm 13 \\ \mathrm{The} \mathrm{length} \mathrm{cannot} \mathrm{be} \mathrm{negative}. \\ \therefore \mathrm{ED}=13 \mathrm{m} \end{gathered}$$

 

Question 1:

$$\begin{gathered} \left(d\right) 5\sqrt{2} \mathrm{cm} \\ \begin{array}{l}\text{ In right angled isoceles triangle, right angled at C, }AC \mathrm{is} \mathrm{equal} \mathrm{to} BC \mathrm{and} AB \text{is the} hypotenuse.\\ \begin{array}{l}\\ \begin{array}{l}{\text{AB}}^2={\text{AC}}^2+{\text{BC}}^2\\ {\text{ =5}}^2+5^2\\ \text{ =50}\\ \therefore \text{ AB=}\sqrt{2\times 25}=5\sqrt{2}\text{ cm}\end{array}\\ \\ \end{array}\end{array} \end{gathered}$$