CBSE Class 10 Mathematics Triangles Worksheet Set 01

Assertion-Reason Questions

The following questions consist of two statements—Assertion(A) and Reason(R). Answer these questions selecting the appropriate option given below:
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true but R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.

 

Question. Assertion (A) : In \( \Delta ABC, DE \parallel BC \) such that \( AD = (7x - 4) \text{ cm}, AE = (5x - 2) \text{ cm}, DB = (3x + 4) \text{ cm} \) and \( EC = 3x \text{ cm} \) then \( x \) equal to 5.
Reason (R) : If a line is drawn parallel to one side of a triangle to intersect the other two sides in different point, than the other two sides are divided in the same ratio.
Answer: Solution : In triangle \( ABC, DE \parallel BC \) so, we have \( \frac{AD}{DB} = \frac{AE}{EC} \)

\( \implies \frac{7x - 4}{3x + 4} = \frac{5x - 2}{3x} \)

\( \implies 21x^2 - 12x = 15x^2 + 20x - 6x - 8 \)

\( \implies 6x^2 - 26x + 8 = 0 \)

\( \implies 3x^2 - 13x + 4 = 0 \)

\( \implies 3x^2 - 12x - x + 4 = 0 \)

\( \implies 3x(x - 4) - 1(x - 4) = 0 \)

\( \implies (x - 4)(3x - 1) = 0 \)

\( \implies x = 4, \frac{1}{3} \)
So, A is false but R is true.
Hence, option (d) is correct.

 

Question. Assertion (A) : If \( \Delta ABC \) is an isosceles triangle right angled at \( C \), then \( AB^2 = 2AC^2 \).
Reason (R) : In right \( \Delta ABC \), right angled at \( C, AB^2 = AC^2 + BC^2 \).
Answer: Solution : In an isosceles \( \Delta ABC \), right angled at \( C \) is \( AB^2 = AC^2 + BC^2 \).

\( \implies AB^2 = AC^2 + AC^2 \)

\( \implies AB^2 = 2AC^2 \quad (\because AC = BC) \)
So, both A and R are true and R is the correct explanation for A.
Hence, option (a) is correct.

 

Question. Assertion (A) : In the \( \Delta ABC, AB = 24 \text{ cm}, BC = 10 \text{ cm} \) and \( AC = 26 \text{ cm} \), then \( \Delta ABC \) is a right angle triangle.
Reason (R) : If in two triangles, their corresponding angles are equal, then the triangles are similar.
Answer: Solution : In the triangle \( ABC \), we have, \( AB^2 + BC^2 = (24)^2 + (10)^2 \)
\( = 576 + 100 = 676 = (26)^2 = AC^2 \)

\( \implies AB^2 + BC^2 = AC^2 \)
\( \therefore ABC \text{ is a right angled triangle.} \)
Also, two triangles are similar if their corresponding angles are equal.
So, both A and R are true but R is not the correct explanation for A.
Hence, option (b) is correct.

 

Question. Assertion (A) : Two similar triangles are always congruent.
Reason (R) : If the areas of two similar triangles are equal then the triangles are congruent.
Answer: Solution : Two similar triangles are congruent only if their areas are equal.
So, A is false but R is true.
Hence, option (d) is correct.

Question. Circles with equal radii are _________ .
Ans. Congruent

Question. In ΔABC, AB = 6√3, AC = 12 cm and BC = 6 cm, then ∠B = _______ .
Ans. 90°

Question. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the _________ ratio.
Ans. Same

Question. The mid-point theorem can be proved by Basic Proprotionality Theorem.
Ans. True

Question. All the similar figures are always congruent.
Ans. False

Question. If the three sides of a triangle are a, √3a and √2a , then the measure of hte angle opposite to longest side is
(a) 45°
(b) 30°
(c) 60°
(d) 90°
Ans. (d) 90°

Question. In the following figure, QA ⊥ AB and PB ⊥ AB, then AQ is

(a) 15 units
(b) 8 units
(c) 5 units
(d) 9 units
Ans. (a) 15 units

Question. The areas of two similar triangles are 144 cm² and 81 cm². If one median of the first triangle is 16 cm, length of corresponding median of the second triangle is
(a) 9 cm
(b) 27 cm
(c) 12 cm
(d) 16 cm
Ans. (c) 12 cm

Question. In the following figure, XY || QR and PX/XQ = PY/YR = 1/2, then

(a) XY = QR
(b) XY = (1/3)QR
(c) XY2 = QR2
(d) XY = (1/2) QR
Ans. (b) XY = (1/3)QR

Question. If ΔABC ~ ΔDEF, BC = 3EF and ar (ΔABC) = 117cm² find area (ΔDEF).
Ans. ar(ABC)/ar(DEF) = (BC/EF)² = (3EF/EF)² = (3/1)²
117 / ar(DEF) = 9 ⇒ ar(DEF) = 13 cm²

Question. Is the triangle with sides 12 cm, 16 cm and 18 cm a right triangle?
Ans. No, because (12)² + (16)² = (18)²

Question. In the given Fig., ∠M = ∠N = 46°, Express x in terms of a, b and c.

Ans. ΔKPN ~ ΔKLM
x/a = c/(b+c)
x = ac/(b+c)

Question. If the ratio of the corresponding sides of two similar triangles is 2 : 3, then find the ratio of their corresponding attitudes.
Ans. 2 : 3

Question. Write the statement of pythagoras theorem.
Ans. In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Question. In the given Fig., D and E are points on sides AB and CA of ΔABC such that ΔB = ∠AED. Show that ΔABC ~ ΔAED.

Ans. ∠B = ∠AED (Given)
∠A = ∠A (Common)
∴ ΔABC ~ ΔAED [AA similarity criterion]

Question. In the given Fig. PQ = 24 cm, QR = 26 cm, ∠PAR = 90°, PA = 6 cm and AR = 8 cm, find ∠QPR.

Ans. PR = √((6)²+(8)²) = 10 cm.
As QR² = PQ² + PR² , therefore ∠QPR = 90°.
(By Converse of Pythagoras theorem)

Question. In the given fig., AB || DC and diagonals AC and BD intersects at O. If OA = 3x – 1 and OB = 2x + 1, OC = 5x – 3 and OD = 6x – 5, find the value of x.

Ans. Draw EO || AB, DE/EA =DO/OB (In ΔADB) and DE/EA = OC/OA (In ΔACD)

(3x-1)/(5x-3) = (2x+1)/(6x-5) ⇒ x = 1/2 or 2
But x = 1/2 is neglected because (5x – 3) get negative value.
So, x = 2 is the required value.

Question. In the adjoining figure ΔABC and ΔDBC are on the same base BC. AD and BC intersect at O. Prove that
area(ΔABC) / area(ΔDBC) = AO/DO.

Ans. Draw AX ⊥ BC and DY ⊥ BC
ar(ΔABC) / ar(ΔDBC) = ((1/2)×BC×AX) / ((1/2)×BC×DY) = AX/DY    ..(1)

ΔAXO ~ ΔDYO         [AA similarity criterion]
AX/DY = AO/DO      ...(2) (C.P.S.T.)
From (1) and (2), we get
ar(ΔABC)/ar(ΔDBC) = AO/DO

Question. Hypotenuse of a right triangle is 25 cm and out of the remaining two sides, one is larger than the other by 5 cm, find the length of the other two sides.
Ans. Let sides of right angled triangle other than hypotenuse be x cm and (x + 5) cm.
By Pythagoras theorem,
(x)² + (x + 5)² = (25)²
x = 15 or – 20
But side is always positive, So, x = 15.
∴ Length of two sides is 15 cm and 20 cm.

Question. In ΔABC, ∠ACB = 90° and CD ⊥ AB. Prove that BC²/AC² = BD/AD.
Ans. 

ΔABC ~ ΔCBD
∴ BC² = AB.BD             ...(1)
ΔABC ~ ΔACD
∴ AC² = AB.AD             ...(2)
Divide (1) by (2), we get
BC²/AC² = BD/AD

Question. In the figure, a point O inside ΔABC is joined to its vertices. From a point D on AO, DE is drawn parallel to AB and from a point E on BO, EF is drawn parallel to BC. Prove that DF || AC.

Ans.

Question. If AD and PS are medians of ΔABC and ΔPQR respectively where ΔABC ~ ΔPQR, Prove that AB/PQ = AD/PS.
Ans. 

Question. In an Obtuse ΔABC (∠B is obtuse), AD is perpendicular to CB produced. Then prove that AC² = AB² + BC² + 2BC × BD.
Ans. 

In ΔADB
AB² = AD² + DB²      ...(i)
In ΔADC
AC² = AD² + DC²
= AD² + (DB + BC)²
= AD² + DB² + BC² + 2BC.BD
= AB² + BC² + 2BC.BD (Using (i))

Question. In the given figure, DE || AC. Which of the following is correct?
x = (a+b)/ay or x = ay/(a+b)

Ans. ΔBED ~ ΔBCA
x/y = a/(a+b)
⇒ x = ay/(a+b)

Question. In an equilateral ΔABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9A D² = 7AB².
Ans. 

 

 

 

Please click on below link to download CBSE Class 10 Mathematics Triangles Worksheet Set A