Solution: False According to the given question, Let’s assume that, $A\text{ }=\text{ }25\text{ }cm$ $B\text{ }=\text{ }5\text{ }cm$ $C\text{ }=\text{ }24\text{ }cm$ Using the Pythagoras Theorem, We...
Solution: We all know that, When the three angles of a triangle are added then their sum equals to 180°. Then, from triangle PQR, $\angle P\text{ }+~\angle Q\text{ }+~\angle R\text{ }=\text{...
Solution: True In triangles PBC and PDE, ∠EPD = ∠BPC [ as vertically opposite angles] $PB/PD\text{ }=\text{ }5/10\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}$… (i) $PC/PE\text{ }=\text{...
Solution: True According to the given question, $PQ\text{ }=\text{ }12.5\text{ }cm$ $PA\text{ }=\text{ }5\text{ }cm$ $BR\text{ }=\text{ }6\text{ }cm$ $PB\text{ }=\text{ }4\text{ }cm$ Then, $QA\text{...
Solution: (d) 100° Explanation: From triangles APB and CPD, $\angle APB\text{ }=~\angle CPD\text{ }=\text{ }50{}^\circ $ (as they are vertically opposite angles) $AP/PD\text{ }=\text{ }6/5$ … (i)...
Solution: (a)Δ PQR~Δ CAB Explanation: From triangles ABC and PQR, we have, AB/QR = BC/PR = CA/PQ When the sides of one triangle are proportional to the sides of the other given triangle, and even...
Solution: (c) BC · DE = AB · EF Explanation: We all know that, If the sides of one triangle are proportionate to the sides of the other triangle, and the corresponding angles are all equal, the...
Solution: (b) 10 cm Explanation: We all know that, A rhombus is a simple quadrilateral with four equal-length sides and diagonals that are perpendicular bisector of each other. Now according to the...
Solution: Given: Δ ABC and AD bisects ∠A, meeting side BC at D. \[AB\text{ }=\text{ }5.6\text{ }cm,\] \[BC\text{ }=\text{ }6\text{ }cm,\] and \[BD\text{ }=\text{ }3.2\text{ }cm\]....
Solution: Let ABCD be a rhombus and AC and BD be the diagonals of ABCD. So, AC = \[24cm\text{ }and\text{ }BD\text{ }=\text{ }10cm\] \[\] We know that diagonals of a rhombus bisect each other at...
(i) AD = a\[\surd \mathbf{3}\] (ii) Area (∆ABC) = \[\surd \mathbf{3}\text{ }{{\mathbf{a}}^{\mathbf{2}}}\] Solution: (i) In ∆ABD and ∆ACD, we have∠ADB = ∠ADC = \[{{90}^{o}}\]AB = AC [Given]AD =...
\[\mathbf{5cm},\]find BC. Solution: Given, An isosceles triangle ABC, AB = AC = \[13cm,\text{ }AD\text{ }=\text{ }5cm\] Required to find: BC In ∆ ADB, by using Pythagoras theorem, we have...
Solution: Given, ABCD is a square. And, F is the mid-point of AB. BE is one third of BC. Area of ∆ FBE = \[108c{{m}^{2}}\] Required to find: length of AC Let’s assume the sides of the square to be...
Solution: From the fig. \[AB\text{ }=\text{ }5cm,\text{ }BC\text{ }=\text{ }12\text{ }cm\text{ }and\text{ }AC\text{ }=\text{ }13\text{ }cm.\] Then, \[A{{C}^{2}}~=\text{ }A{{B}^{2}}~+\text{...
Solution: We have, In ∆BAC, by Pythagoras theorem, we have \[\begin{array}{*{35}{l}} B{{C}^{2}}~=\text{ }A{{B}^{2}}~+\text{ }A{{C}^{2}} \\ \Rightarrow...
Solution: Comparing with the figure, it’s given that AC = \[14\text{ }m,\text{ }DC\text{ }=\text{ }12m\text{ }and\text{ }ED\text{ }=\text{ }BC\text{ }=\text{ }9\text{ }m\] Construction: Draw...
Solution: Let’s assume the length of ladder to be, AD = BE = x m So, in ∆ACD, by Pythagoras theorem We have, \[\begin{array}{*{35}{l}} ...
Solution: Given, ∆ABC, AB = AC = \[25\text{ }cm\text{ }and\text{ }BC\text{ }=\text{ }14.\] \[\] In ∆ABD and ∆ACD, we see...
Solution: Let CD and AB be the poles of height \[11m\text{ }and\text{ }6m.\] Then, its seen that CP = \[11\text{ }\text{ }6\text{ }=\text{ }5m.\] From the figure, AP should be \[12m\] (given) In...
Solution: In ∆ABC, by Pythagoras theorem \[\begin{array}{*{35}{l}} A{{B}^{2}}~+\text{ }B{{C}^{2}}~=\text{ }A{{C}^{2}} \\ \Rightarrow {{15}^{2}}~+\text{...
Solution: ...
\[\begin{array}{*{35}{l}} \left( \mathbf{i} \right)\text{ }\mathbf{a}\text{ }=\text{ }\mathbf{7}\text{ }\mathbf{cm},\text{ }\mathbf{b}\text{ }=\text{ }\mathbf{24}\text{...
Solution: We have, Sides of triangle as \[\begin{array}{*{35}{l}} AB\text{ }=\text{ }3\text{ }cm \\ BC\text{ }=\text{ }4\text{ }cm \\ AC\text{...