SSC (English Medium) १० वीं कक्षा - Maharashtra State Board Important Questions for Geometry Mathematics 2

From given figure, In ∆ABC, AB ⊥ BC, AB = BC 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

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

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

Complete the following activity to find the length of hypotenuse of right angled triangle, if sides of right angle are 9 cm and 12 cm.

Activity: In ∆PQR, m∠PQR = 90°

By Pythagoras Theorem,

PQ² + `square` = PR²    ......(I)

∴ PR² = 9² + 12² 

∴ PR² = `square` + 144

∴ PR² = `square`

∴ PR = 15

∴ Length of hypotenuse of triangle PQR is `square` cm.

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.

A congruent side of an isosceles right angled triangle is 7 cm, Find its perimeter

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

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

In ΔABC, ∠ABC = 90°, ∠BAC = ∠BCA = 45°. If AC = `9sqrt(2)`, then find the value of AB.

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`

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

If a and b are natural numbers and a > b If (a² + b²), (a² – b²) and 2ab are the sides of the triangle, then prove that the triangle is right-angled. Find out two Pythagorean triplets by taking suitable values of a and b.

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`

ΔPQR, is a right angled triangle with ∠Q = 90°, QR = b, and A(ΔPQR) = a. If QN ⊥ PR, then prove that QN = `(2ab)/sqrt(b^4 + 4a^2)`

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.

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.

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).