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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Additional Questions

Question 1.

In figure if PQ || RS, Prove that ∆POQ ~ ∆SOR

Solution:

PQ || RS

So, ∠P = ∠S (A Hernate angles)

and ∠Q = ∠R

Also, ∠POQ = ∠SOR (vertically opposite angle)

∴ ∆POQ ~ ∆SOR (AAA similarity criterion)

Question 2.

In figure OA . OB = OC . OD Show that ∠A = ∠C and ∠B = ∠D

Solution:

OA . OB = OC . OD (Given)

Also we have ∠AOD = ∠COB

(vertically opposite angles) …………. (2)

From (1) and (2)

∴ ∆AOD ~ ∆COB (SAS similarity criterion)

So, ∠A = ∠C and ∠B = ∠D

(corresponding angles of similar triangles)

Question 3.

In figure the line segment XY is parallel to side AC of ∆ABC and it divides the triangle into two parts of equal areas. Find the ratio \(\frac{\mathbf{A X}}{\mathbf{A B}}\)

Solution:

Given XY||AC

Question 4.

In AD⊥BC, prove that AB^{2} + CD^{2} = BD^{2} + AC^{2}.

Solution:

From ∆ADC, we have

AC^{2} = AD^{2} + CD^{2} …………… (1)

(Pythagoras theorem)

From ∆ADB, we have

AB^{2} = AD^{2} + BD^{2} …………. (2)

(Pythagoras theorem)

Subtracting (1) from (2) we have,

AB^{2} – AC^{2} = BD^{2} – CD^{2}

AB^{2} + CD^{2} = BD^{2} + AC^{2}

Question 5.

BL and CM are medians of a triangle ABC right angled at A.

Prove that 4(BL^{2} + CM^{2}) = 5BC^{2}.

Solution:

BL and CM are medians at the ∆ABC in which

A = ∠90°.

From ∆ABC

BC^{2} = AB^{2} + AC^{2} …………… (1)

(Pythagoras theorem)

4CM^{2} = 4AC^{2} + AB^{2} …………… (3)

Adding (2) and (3), we have

4(BL^{2} + CM^{2}) = 5(AC^{2} + AB^{2})

4(BL^{2} + CM^{2}) = 5BC^{2} [From (1)]

Question 6.

Prove that in a right triangle, the square of the hypotenure is equal to the sum of the squares of the others two sides.

Solution:

Proof:

We are given a right triangle ABC right angled at B.

We need to prove that AC^{2} = AB^{2} + BC^{2}

Let us draw BD ⊥ AC

Now, ∆ADB ~ ∆ABC

\(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AB}}{\mathrm{AC}}\) (sides are proportional)

AD . AC = AB^{2} …………. (1)

Also, ∆BDC ~ ∆ABC

\(\frac{C D}{B C}=\frac{B C}{A C}\)

CD . AC = BC^{2} …………. (2)

Adding (1) and (2)

AD . AC + CD . AC = AB^{2} + BC^{2}

AC(AD + CD) = AB^{2} + BC^{2}

AC . AC = AB^{2} + BC^{2}

AC^{2} = AB^{2} + BC^{2}

Question 7.

A ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall and its top reaches a window 6 m above the ground. Find the length of the ladder.

Solution:

Let AB be the ladder and CA be the wall with the window at A.

Also, BC = 2.5 m and CA = 6 m

From Pythagoras theorem,

AB^{2} = BC^{2} + CA^{2}

= (2.5)^{2} + (6)^{2}

= 42.25

AB = 6.5

Thus, length at the ladder is 6.5 m.

Question 8.

In figure O is any point inside a rectangle ABCD. Prove that OB^{2} + OD^{2} = OA^{2} + OC^{2}.

Solution:

Through O, draw PQ||BC so that P lies on AB and Q lies on DC.

Now, PQ||BC

PQ⊥AB and PQ⊥DC (∵ ∠B = 90° and ∠C = 90°)

So, ∠BPQ = 90° and ∠CQP = 90°

Therefore BPQC and APQD are both rectangles.

Now from ∆OPB,

OB^{2} = BP^{2} + OP^{2} ……………. (1)

Similarly from ∆OQD,

OD^{2} = OQ^{2} + DQ^{2} …………. (2)

From ∆OQC, we have

OC^{2} = OQ^{2} + CQ^{2} ……….. (3)

∆OAP, we have

OA^{2} = AP^{2} + OP^{2} …………. (4)

Adding (1) and (2)

OB^{2} + OD^{2} = BP^{2} + OP^{2} + OQ^{2} + DQ^{2} (As BP = CQ and DQ = AP)

= CQ^{2} + OP^{2} + OQ^{2} + AP^{2}

= CQ^{2} + OQ^{2} + OP^{2} + AP^{2}

= OC^{2} + OA^{2} [From (3) and (4)]

Question 9.

In ∠ACD = 90° and CD⊥AB. Prove that \(\frac{\mathbf{B C}^{2}}{\mathbf{A C}^{2}}=\frac{\mathbf{B D}}{\mathbf{A D}}\)

Solution:

∆ACD ~ ∆ABC

So,

Question 10.

The perpendicular from A on side BC at a ∆ABC intersects BC at D such that DB = 3 CD. Prove that 2AB^{2} = 2AC^{2} + BC^{2}.

Solution:

We have DB = 3 CD

Since ∆ABD is a right triangle (i) right angled at D

AB^{2} – AD^{2} + BD^{2} …………. (ii)

By ∆ACD is a right triangle right angled at D

AC^{2} = AD^{2} + CD^{2} ………… (iii)

Subtracting equation (iii) from equation (ii),

we got