SSC (English Medium) 10th Standard - Maharashtra State Board Question Bank Solutions for Geometry Mathematics 2

In the figure, ΔPQR is right angled at Q, seg QS ⊥ seg PR. Find x, y.

The perimeter of an arc of radius 4.2 cm is 12.8 cm. Determine the angle subtended by the arc at the centre of circle.

In the following figure, XY = 10 cm and LT = 4 cm. Find the length of XT.

A circle touches side BC at point P of the ΔABC, from outside of the triangle. Further extended lines AC and AB are tangents to the circle at N and M respectively. Prove that : AM = `1/2` (Perimeter of ΔABC)

In the given figure, m(arc NS) = 125°, m(arc EF) = 37°, find the measure ∠NMS.

In the above figure `square`ABCD is a rectangle. If AB = 5, AC = 13, then complete the following activity to find BC.

Activity: ΔABC is a `square` triangle.

∴ By Pythagoras theorem

AB² + BC² = AC²

∴ 25 + BC² = `square`

∴ BC² = `square`

∴ BC = `square`

Construct two concentric circles with centre O with radii 3 cm and 5 cm. Construct a tangent to a smaller circle from any point A on the larger circle. Measure and write the length of the tangent segment. Calculate the length of the tangent segment using Pythagoras' theorem.

In the right-angled triangle ABC, Hypotenuse AC = 10 and side AB = 5, then what is the measure of ∠A?

If tan θ = `12/5`, then 5 sin θ – 12 cos θ = ?

From the information in the figure, complete the following activity to find the length of the hypotenuse AC.

AB = BC = `square`

∴ ∠BAC = `square`

Side opposite angle 45° = `square/square` × Hypotenuse

∴ `5sqrt(2) = 1/square` × AC

∴ AC = `5sqrt(2) xx square = square`

AB, BC and AC are three sides of a right-angled triangle having lengths 6 cm, 8 cm and 10 cm, respectively. To verify the Pythagoras theorem for this triangle, fill in the boxes:

ΔABC is a right-angled triangle and ∠ABC = 90°.

So, by the Pythagoras theorem,

`square` + `square` = `square`

Substituting 6 cm for AB and 8 cm for BC in L.H.S.

`square` + `square` = `square` + `square`

= `square` + `square`

= `square`

Substituting 10 cm for AC in R.H.S.

`square` = `square`

= `square`

Since, L.H.S. = R.H.S.

Hence, the Pythagoras theorem is verified.

There is a ladder of length 32 m which rests on a pole. If the height of pole is 18 m, determine the distance between the foot of ladder and the pole.

A person starts his trip from home. He moves 24 km in south direction and then starts moving towards east. He travels 7 km in that direction and finally reaches his destination. How far is the destination from his home?

Find the side of a square whose diagonal is `10sqrt2` cm.

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`

Find the side and perimeter of a square whose diagonal is `13sqrt2` cm. 

From given figure, In ∆ABC, AB ⊥ BC, AB = BC then m∠A = ?

From given figure, In ∆ABC, AB ⊥ BC, AB = BC, AC = `2sqrt(2)` then `l` (AB) = ?

From given figure, In ∆ABC, AB ⊥ BC, AB = BC, AC = `5sqrt(2)`, then what is the height of ∆ABC?