SSC (English Medium) इयत्ता १० वी - Maharashtra State Board Question Bank Solutions

If m and n are real numbers and m > n, if m² + n², m² – n² and 2 mn are the sides of the triangle, then prove that the triangle is right-angled. (Use the converse of the Pythagoras theorem). Find out two Pythagorian triplets using convenient values of m and n.

If ΔABC ∼ ΔDEF, length of side AB is 9 cm and length of side DE is 12 cm, then find the ratio of their corresponding areas.

What is the distance of the point (– 5, 4) from the origin?

In the figure, PQ ⊥ BC, AD ⊥ BC. To find the ratio of A(ΔPQB) and A(ΔPBC), complete the following activity.

Given: PQ ⊥ BC, AD ⊥ BC

Now, A(ΔPQB)  = `1/2 xx square xx square`

A(ΔPBC)  = `1/2 xx square xx square`

Therefore, 

`(A(ΔPQB))/(A(ΔPBC)) = (1/2 xx square xx square)/(1/2 xx square xx square)`

= `square/square`

Find the value of y, if the points A(3, 4), B(6, y) and C(7, 8) are collinear.

Draw a line segment AB of length 10 cm and divide it internally in the ratio of 2:5 Justify the division of line segment AB.

Find the distance between the points O(0, 0) and P(3, 4).

Show that points A(–1, –1), B(0, 1), C(1, 3) are collinear.

In the following figure, in ΔABC, ∠B = 90°, ∠C = 60°, ∠A = 30°, AC = 18 cm. Find BC.

Find the height of an equilateral triangle whose side is 6 units.

In the given figure. Find RP and PS using the information given in ∆PSR.

In the given figure, ∆PQR is an equilateral triangle. Point S is on seg QR such that QS = n\[\frac{1}{3}\] QR.

Prove that: 9 PS² = 7 PQ²

In right-angled triangle PQR, if ∠P = 60°, ∠R = 30° and PR = 12, then find the values of PQ and QR. 

In ΔDEF, if ∠E = 90°, then find the value of ∠D + ∠F. 

Choose the correct alternative:

Out of given triplets, which is a Pythagoras triplet?

Choose the correct alternative:

The diagonal of a square is `10sqrt(2)` cm then its perimeter is ______

From given figure, In ∆ABC, If ∠ABC = 90° ∠CAB=30°, AC = 14 then for finding value of AB and BC, complete the following activity.

Activity: In ∆ABC, If ∠ABC = 90°, ∠CAB=30°

∴ ∠BCA = `square`

By theorem of 30° – 60° – 90° triangle,

∴ `square = 1/2` AC and `square = sqrt(3)/2` AC

∴ BC = `1/2 xx square` and AB = `sqrt(3)/2 xx 14`

∴ BC = 7 and AB = `7sqrt(3)`

From given figure, in ∆MNK, if ∠MNK = 90°, ∠M = 45°, MK = 6, then for finding value of MN and KN, complete the following activity.

Activity:

In ∆MNK, ∠MNK = 90°, ∠M = 45°  …...[Given]

∴ ∠K = `square`      .....[Remaining angle of ∆MNK]

By theorem of 45° – 45° – 90° triangle,

∴ `square = 1/sqrt(2)` MK and `square = 1/sqrt(2)` MK

∴ MN = `1/sqrt(2) xx square` and KN = `1/sqrt(2) xx 6`

∴ MN = `3sqrt(2)` and KN = `3sqrt(2)`

From the given figure, in ∆ABC, if AD ⊥ BC, ∠C = 45°, AC = `8sqrt(2)` , BD = 5, then for finding value of AD and BC, complete the following activity.

Activity: In ∆ADC, if ∠ADC = 90°, ∠C = 45°    ......[Given]

∴ ∠DAC = `square`   .....[Remaining angle of ∆ADC]

By theorem of 45° – 45° – 90° triangle,

∴ `square = 1/sqrt(2)` AC and `square = 1/sqrt(2)` AC

∴ AD =`1/sqrt(2) xx square` and DC = `1/sqrt(2) xx 8sqrt(2)`

∴ AD = 8 and DC = 8

∴ BC = BD +DC

= 5 + 8

= 13

As shown in figure, LK = `6sqrt(2)` then

1. MK = ?
2. ML = ?
3. MN = ?