SSC (English Medium) 10th Standard - Maharashtra State Board Question Bank Solutions

If the length of an arc of the sector of a circle is 20 cm and if the radius is 7 cm, find the area of the sector. 

If sinθ = `8/17`, where θ is an acute angle, find the value of cos θ by using identities.

Show that: 

`sqrt((1-cos"A")/(1+cos"A"))=cos"ecA - cotA"`

In ΔPQR, ∠P = 30°, ∠Q = 60°, ∠R = 90° and PQ = 12 cm, then find PR and QR.

ΔAMT∼ΔAHE, construct Δ AMT such that MA = 6.3 cm, ∠MAT=120°, AT = 4.9 cm and `"MA"/"HA"=7/5`, then construct ΔAHE.

Choose the correct alternative:

In right angled triangle, if sum of the squares of the sides of right angle is 169, then what is the length of the hypotenuse?

Choose the correct alternative:

A rectangle having length of a side is 12 and length of diagonal is 20, then what is length of other side?

Choose the correct alternative:

If the length of diagonal of square is √2, then what is the length of each side?

Choose the correct alternative:

If length of both diagonals of rhombus are 60 and 80, then what is the length of side?

Choose the correct alternative:

In ∆ABC, AB = `6sqrt(3)` cm, AC = 12 cm, and BC = 6 cm, then m∠A = ?

If a triangle having sides 50 cm, 14 cm and 48 cm, then state whether given triangle is right angled triangle or not

If a triangle having sides 8 cm, 15 cm and 17 cm, then state whether given triangle is right angled triangle or not

A rectangle having dimensions 35 m × 12 m, then what is the length of its diagonal?

In ∆LMN, l = 5, m = 13, n = 12 then complete the activity to show that whether the given triangle is right angled triangle or not.
*(l, m, n are opposite sides of ∠L, ∠M, ∠N respectively)

Activity: In ∆LMN, l = 5, m = 13, n = `square`

∴ l² = `square`, m² = 169, n² = 144.

∴ l² + n² = 25 + 144 = `square`

∴ `square` + l² = m²

∴By Converse of Pythagoras theorem, ∆LMN is right angled triangle.

In ΔABC, AB = 9 cm, BC = 40 cm, AC = 41 cm. State whether ΔABC is a right-angled triangle or not. Write reason.

Prove that: (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ

Proof: L.H.S. = (sec θ – cos θ) (cot θ + tan θ)

= `(1/square - cos θ) (square/square + square/square)` ......`[∵ sec θ = 1/square, cot θ = square/square and tan θ = square/square]`

= `((1 - square)/square) ((square + square)/(square  square))`

= `square/square xx 1/(square  square)`  ......`[(∵ square + square = 1),(∴ square = 1 - square)]`

 = `square/(square  square)`

= tan θ.sec θ

= R.H.S.

∴ L.H.S. = R.H.S.

∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ

In the given figure, triangle PQR is right-angled at Q. S is the mid-point of side QR. Prove that QR² = 4(PS² – PQ²).

In a right angled triangle, right-angled at B, lengths of sides AB and AC are 5 cm and 13 cm, respectively. What will be the length of side BC?

In the figure given above, `square`ABCD is a square and a circle is inscribed in it. All sides of a square touch the circle. If AB = 14 cm, find the area of shaded region.

Solution:

Area of square = `(square)^2`   ......(Formula)

= 14²

= `square  "cm"^2`

Area of circle = `square`    ......(Formula)

= `22/7 xx 7 xx 7`

= 154 cm²

(Area of shaded portion) = (Area of square) - (Area of circle)

= 196 − 154

= `square  "cm"^2`

In the given figure, altitudes YZ and XT of ∆WXY intersect at P. Prove that,

1. `square`WZPT is cyclic.
2. Points X, Z, T, Y are concyclic.