List of top Mathematics Questions asked in Maharashtra Class X Board

In the following figure \(\triangle\) ABC, B-D-C and BD = 7, BC = 20, then find \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\). 

In the following figure \(\angle\)MNP = 90\(^\circ\), seg NQ \(\perp\) seg MP, MQ = 9, QP = 4, find NQ. 

The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\)) 

Activity : 
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14. 
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.
 

In the following figure chord MN and chord RS intersect at point D. If RD = 15, DS = 4, MD = 8, find DN by completing the following activity: 

Activity : 
\(\therefore\) MD \(\times\) DN = \(\boxed{\phantom{SD}}\) \(\times\) DS \(\dots\) (Theorem of internal division of chords)
\(\therefore\) \(\boxed{\phantom{8}}\) \(\times\) DN = 15 \(\times\) 4
\(\therefore\) DN = \(\frac{\boxed{\phantom{60}}}{8}\)
\(\therefore\) DN = \(\boxed{\phantom{7.5}}\)

An observer at a distance of 10 m from tree looks at the top of the tree, the angle of elevation is 60\(^\circ\). To find the height of tree complete the activity. (\(\sqrt{3} = 1.73\)) 

Activity : 
In the figure given above, AB = h = height of tree, BC = 10 m, distance of the observer from the tree. 
Angle of elevation (\(\theta\)) = \(\angle\)BCA = 60\(^\circ\) 
tan \(\theta\) = \(\frac{\boxed{\phantom{AB}}}{BC}\) \(\dots\) (I)
tan 60\(^\circ\) = \(\boxed{\phantom{\sqrt{3}}}\) \(\dots\) (II)
\(\frac{AB}{BC} = \sqrt{3}\) \(\dots\) (From (I) and (II))
AB = BC \(\times\) \(\sqrt{3}\) = 10\(\sqrt{3}\)
AB = 10 \(\times\) 1.73 = \(\boxed{\phantom{17.3}}\)
\(\therefore\) height of the tree is \(\boxed{\phantom{17.3}}\) m.

Solve the following sub-questions (any four): In \( \triangle ABC \), \( DE \parallel BC \). If \( DB = 5.4 \, \text{cm} \), \( AD = 1.8 \, \text{cm} \), \( EC = 7.2 \, \text{cm} \), then find \( AE \).

In the figure given below, find RS and PS using the information given in \(\triangle\)PSR. 

In the following figure, circle with centre D touches the sides of \(\angle\)ACB at A and B. If \(\angle\)ACB = 52\(^\circ\), find measure of \(\angle\)ADB. 

In the following figure, XY \(||\) seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC.

Activity: 
2AX = 3BX (Given) 
\[\therefore \frac{AX}{BX} = \frac{3}{\boxed{2}} \ \\ \frac{AX + BX}{BX} = \frac{3 + 2}{2} \quad \text{(by componendo)} \ \\ \frac{BA}{BX} = \frac{5}{2} \quad \dots \text{(I)} [6pt] \\ \text{Now } \triangle BCA \sim \triangle BYX ; (\boxed{\text{AA}} \text{ test of similarity}) [4pt] \\ \therefore \frac{BA}{BX} = \frac{AC}{XY} \quad \text{(corresponding sides of similar triangles)} [4pt] \\ \frac{5}{2} = \frac{AC}{9} \quad \text{from (I)} [4pt] \\ \therefore AC = \boxed{22.5}\]
 

Complete the following activity to prove that the sum of squares of diagonals of a rhombus is equal to the sum of the squares of the sides. 

Given: PQRS is a rhombus. Diagonals PR and SQ intersect each other at point T. 
To prove: PS\(^2\) + SR\(^2\) + QR\(^2\) + PQ\(^2\) = PR\(^2\) + QS\(^2\) 
Activity: Diagonals of a rhombus bisect each other. 
In \(\triangle\)PQS, PT is the median and in \(\triangle\)QRS, RT is the median. 
\(\therefore\) by Apollonius theorem, 
 

\[\begin{aligned} PQ^2 + PS^2 &= \boxed{\phantom{X}} + 2QT^2 \quad \dots \text{(I)} \\ QR^2 + SR^2 &= \boxed{\phantom{X}} + 2QT^2 \quad \dots \text{(II)} \\ \text{Adding (I) and (II),} \quad PQ^2 + PS^2 + QR^2 + SR^2 &= 2(PT^2 + \boxed{\phantom{X}}) + 4QT^2 \\ &= 2(PT^2 + \boxed{\phantom{X}}) + 4QT^2 \quad (\text{RT = PT}) \\ &= 4PT^2 + 4QT^2 \\ &= (\boxed{\phantom{X}})^2 + (2QT)^2 \\ \therefore \quad PQ^2 + PS^2 + QR^2 + SR^2 &= PR^2 + \boxed{\phantom{X}} \\ \end{aligned}\]