SSC (English Medium) १० वीं कक्षा - Maharashtra State Board Question Bank Solutions for Geometry Mathematics 2

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 = ?

In ΔABC, ∠ABC = 90° and ∠ACB = θ. Then write the ratios of sin θ and tan θ from the figure.

Prove that sec θ + tan θ = `cos θ/(1 - sin θ)`.

Proof: L.H.S. = sec θ + tan θ

= `1/square + square/square`

= `square/square`  ......`(∵ sec θ = 1/square, tan θ = square/square)`

= `((1 + sin θ) square)/(cos θ  square)`  ......[Multiplying `square` with the numerator and denominator]

= `(1^2 - square)/(cos θ  square)`

= `square/(cos θ  square)`

= `cos θ/(1 - sin θ)` = R.H.S.

∴ L.H.S. = R.H.S.

∴ sec θ + tan θ = `cos θ/(1 - sin θ)`

Find the value of sin 0° + cos 0° + tan 0° + sec 0°.

Find the value of sin 45° + cos 45° + tan 45°.

What will be the value of sin 45° + `1/sqrt(2)`?

In the given figure, if sin θ = `7/13`, which angle will be θ?

Prove that: cot θ + tan θ = cosec θ·sec θ

Proof: L.H.S. = cot θ + tan θ

= `square/square + square/square`  ......`[∵ cot θ = square/square, tan θ = square/square]`

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

= `1/(square xx square)`

= `1/square xx 1/square`

= cosec θ·sec θ  ......`[∵ "cosec"  θ = 1/square, sec θ = 1/square]`

= R.H.S.

∴ L.H.S. = R.H.S.

∴ cot θ + tan θ = cosec·sec θ

If sec θ = `1/2`, what will be the value of cos θ?

Find will be the value of cos 90° + sin 90°.

Write down the equation of a line whose slope is 3/2 and which passes through point P, where P divides the line segment AB joining A(-2, 6) and B(3, -4) in the ratio 2 : 3.

ΔRST ~ ΔUAY, In ΔRST, RS = 6 cm, ∠S = 50°, ST = 7.5 cm. The corresponding sides of ΔRST and ΔUAY are in the ratio 5 : 4. Construct ΔUAY.

Construct the circumcircle and incircle of an equilateral triangle ABC with side 6 cm and centre O. Find the ratio of radii of circumcircle and incircle.

Construct the circumcircle and incircle of an equilateral ∆XYZ with side 6.5 cm and centre O. Find the ratio of the radii of incircle and circumcircle.