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

Prove that, in a right angled triangle, the square of the hypotenuse is
equal to the sum of the squares of remaining two sides.

In right angled triangle PQR,
if ∠ Q = 90°, PR = 5,
QR = 4 then find PQ and hence find tan R.

Choose the correct alternative: 

ΔABC and ΔDEF are equilateral triangles. If ar(ΔABC): ar(ΔDEF) = 1 : 2 and AB = 4, then what is the length of DE?

In the above figure, line l || line m and line n is a transversal. Using the given information find the value of x. 

Choose the correct alternative: 
sinθ × cosecθ =?

In the above figure, line AB || line CD || line EF, line l, and line m are its transversals. If  AC = 6, CE = 9. BD = 8, then complete the following activity to find DF. 

Activity :

`"AC"/"" = ""/"DF"`   (Property of three parallel lines and their transversal) 

∴ `6/9 = ""/"DF"`

∴ `"DF"  = "___"`

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.

From given figure, In ∆ABC, AD ⊥ BC, then prove that AB² + CD² = BD² + AC² by completing activity.

Activity: From given figure, In ∆ACD, by pythagoras theorem

AC² = AD² + `square`

∴ AD² = AC² – CD²   ...(I)

Also, In ∆ABD, by pythagoras theorem,

AB² = `square` + BD²

∴ AD² = AB² – BD²   ...(II)

∴ `square` – BD² = AC² – `square`

∴ AB² + CD² = AC² + BD²

A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of wall. Complete the given activity.

Activity: As shown in figure suppose

PR is the length of ladder = 10 m

At P – window, At Q – base of wall, At R – foot of ladder

∴ PQ = 8 m

∴ QR = ?

In ∆PQR, m∠PQR = 90°

By Pythagoras Theorem,

∴ PQ² + `square` = PR²    ...(I)

Here, PR = 10, PQ = `square`

From equation (I)

8² + QR² = 10²

QR² = 10² – 8²

QR² = 100 – 64

QR² = `square`

QR = 6

∴ The distance of foot of the ladder from the base of wall is 6 m.

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.

From given figure, In ∆PQR, If ∠QPR = 90°, PM ⊥ QR, PM = 10, QM = 8, then for finding the value of QR, complete the following activity.

Activity: In ∆PQR, If ∠QPR = 90°, PM ⊥ QR,   ...(Given)

In ∆PMQ, By Pythagoras Theorem,

∴ PM² + `square` = PQ²   ...(I)

∴ PQ² = 10² + 8² 

∴ PQ² = `square` + 64

∴ PQ² = `square`

∴ PQ = `sqrt(164)`

Here, ∆QPR ~ ∆QMP ~ ∆PMR

∴ ∆QMP ~ ∆PMR

∴ `(PM)/(RM) = (QM)/(PM)`

∴ PM² = RM × QM

∴ 10² = RM × 8

RM = `100/8 = square`

And,

QR = QM + MR

QR = `square` + `25/2 = 41/2`

Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq. cm. Complete the following activity.

Activity: As shown in the figure, LMNT is a rectangle.

∴ Area of rectangle = length × breadth

∴ Area of rectangle = `square` × breadth

∴ 192 = `square` × breadth

∴ Breadth = 12 cm.

Also, ∠TLM = 90°   ...(Each angle of reactangle is right angle)

In ∆TLM, By Pythagoras theorem

∴ TM² = TL² + `square`

∴ TM² = 12² + `square`

∴ TM² = 144 + `square`

∴ TM² = 400

∴ TM = 20

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

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)`

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, ∠MNP = 90°, seg NQ ⊥ seg MP, MQ = 9, QP = 4, find NQ.

In the given figure, ∠QPR = 90°, seg PM ⊥ seg QR and Q–M–R, PM = 10, QM = 8, find QR.