Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Additional Questions

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 AX/AB
Solution:
Given XY||AC

Question 4.
In AD⊥BC, prove that AB² + CD² = BD² + AC².
Solution:
From ∆ADC, we have
AC² = AD² + CD² …………… (1)
(Pythagoras theorem)
From ∆ADB, we have
AB² = AD² + BD² …………. (2)
(Pythagoras theorem)
Subtracting (1) from (2) we have,
AB² – AC² = BD² – CD²
AB² + CD² = BD² + AC²

Question 5.
BL and CM are medians of a triangle ABC right angled at A.
Prove that 4(BL² + CM²) = 5BC².
Solution:
BL and CM are medians at the ∆ABC in which
A = ∠90°.
From ∆ABC
BC² = AB² + AC² …………… (1)
(Pythagoras theorem)

4CM² = 4AC² + AB² …………… (3)
Adding (2) and (3), we have
4(BL² + CM²) = 5(AC² + AB²)
4(BL² + CM²) = 5BC² [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 AC² = AB² + BC²
Let us draw BD ⊥ AC
Now, ∆ADB ~ ∆ABC
AD/AB=AB/AC (sides are proportional)
AD . AC = AB² …………. (1)
Also, ∆BDC ~ ∆ABC
CD/BC=BC/AC
CD . AC = BC² …………. (2)
Adding (1) and (2)
AD . AC + CD . AC = AB² + BC²
AC(AD + CD) = AB² + BC²
AC . AC = AB² + BC²
AC² = AB² + BC²

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,
AB² = BC² + CA²
= (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 OB² + OD² = OA² + OC².
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,
OB² = BP² + OP² ……………. (1)
Similarly from ∆OQD,
OD² = OQ² + DQ² …………. (2)
From ∆OQC, we have
OC² = OQ² + CQ² ……….. (3)
∆OAP, we have
OA² = AP² + OP² …………. (4)
Adding (1) and (2)
OB² + OD² = BP² + OP² + OQ² + DQ² (As BP = CQ and DQ = AP)
= CQ² + OP² + OQ² + AP²
= CQ² + OQ² + OP² + AP²
= OC² + OA² [From (3) and (4)]

Question 9.
In ∠ACD = 90° and CD⊥AB. Prove that BC²/AC²=BD/AD
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 2AB² = 2AC² + BC².
Solution:
We have DB = 3 CD

Since ∆ABD is a right triangle (i) right angled at D
AB² – AD² + BD² …………. (ii)
By ∆ACD is a right triangle right angled at D
AC² = AD² + CD² ………… (iii)
Subtracting equation (iii) from equation (ii),
we got