M.L Aggarwal Solutions

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ML Aggarwal Solution Class 9 Chapter 11 Mid Point Theorem MCQS

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  MCQS Choose the correct answer from the given four options (1 to 6): Question 1 In a ∆ABC, AB = 3 cm, BC = 4 cm and CA = 5 cm. IfD and E are mid-points of AB and BC respectively, then the length of DE is (a) 1.5 cm (b) 2 cm (c) 2.5 cm (d) 3.5 cm Sol : In ∆ABC, D and E are the mid-points of sides AB and BC ∴DE$=\frac{1}{2}$AC But AC=5 cm ∴DE$=\frac{1}{2}\times 5=\frac{5}{2}$ cm =2.5cm Ans (c) Question 2 In the given figure, ABCD is a rectangle in which AB = 6 cm and AD = 8 cm. If P and Q are mid-points of the sides BC and CD respectively, then the length of PQ is (a) 7 cm (b) 5 cm (c) 4 cm (d) 3 cm Sol : In the given figure ABCD is a rectangle AB=6 cm , AD=8 cm P and Q are mid points of BC and CD ∴PQ$=\frac{1}{2}$BD But BD$=\sqrt{BC^2+CD^2}$ (pythagoras theorem) $=\sqrt{8^2+6^2}$ $=\sqrt{64+36}=\sqrt{100}$ =10 cm ∴PQ$=\frac{1}{2}$BD $=\frac{1}{2}\times 10=5$ cm Ans (b) Question 3 D and E are mid-points of the sides AB and AC of ∆ABC and O is any point on the side BC. O is joi...
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ML Aggarwal Solution Class 9 Chapter 13 Rectilinear Figures Exercise 13.1

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 Exercise 13.1 Question 1 If two angles of a quadrilateral are 40° and 110° and the other two are in the ratio 3 : 4, find these angles. Sol : Sum of four angles of a quadrilateral=360° Sum of two given angles =40°+110°=150° ∴Sum of remaining two angles =360°-150°=210° ∴Third angle$=\frac{210^{\circ} \times 3}{3+4}$ $=\frac{210^{\circ} \times 3}{7}=90^{\circ}$ and fourth angle $=\frac{210^{\circ} \times 4}{3+4}$ $=\frac{210^{\circ} \times 4}{7}$ =120° Question 2 If the angles of a quadrilateral, taken in order, are in the ratio 1 : 2 : 3 : 4, prove that it is a trapezium. Sol : In trapezium ABCD ∠A : ∠B : ∠C : ∠D=1 : 2 : 3 : 4 Sum of angles of the quad ABCD=360° Sum of the ratio's=1+2+3+4=10 ∴$ \angle \mathrm{A}=\frac{360^{\circ} \times 1}{10}=36^{\circ}$ $\angle B=\frac{360^{\circ} \times 2}{10}=72^{\circ}$ $\angle C=\frac{360^{\circ} \times 3}{10}=108^{\circ}$ $\angle D=\frac{360^{\circ} \times 4}{10}=144^{\circ}$ Now , ∠A+∠D=36°+114°=180° ∵∠A+∠D=180° and there are co-interi...
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ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.2

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Exercise 15.2 Question 1 If O is the centre of the circle, find the value of x in each of the following figures (using the given information): Sol : From the figure (i) ABCD is a cyclic quadrilateral From the figure $\therefore$ Ext. $\angle \mathrm{DCE}=\angle \mathrm{BAD}$ $\therefore  \angle \mathrm{BAD}=x^{\circ}$ Now arc BD subtends $\angle \mathrm{BOD}$ at the centre and $\angle \mathrm{BAD}$ at the remaining part of the circle. $\therefore \angle B O D=2 \angle B A D=2 x$ $\Rightarrow  2 x=150^{\circ} \Rightarrow x=75^{\circ}$ (ii) $\angle \mathrm{BCD}+\angle \mathrm{DCE}=180^{\circ}$ (Linear pair) Figure to be added $\Rightarrow \angle B C D+80^{\circ}=180^{\circ}$ $\Rightarrow \angle B C D=180^{\circ}-80^{\circ}=100^{\circ}$ ∵Arc BAD subtends reflex ∠BOD at the centre and ∠BCD at the remaining part of the circle $\therefore$ Reflex $\angle \mathrm{BOD}=2 \angle \mathrm{BCD}$ $\Rightarrow x^{0}=2 \times 100^{\circ}=200^{\circ}$ (iii) In $\Delta \mathrm{ACB}$ ...
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