Free Triangles 02 Practice Test - 10th Grade
Question 1
△ABC is such thatAB=3 cm,BC=2 cm and CA=2.5 cm. △DEF is similar to △ABC. If EF=4 cm, then the perimeter of △DEF is ____.
7.5 cm
15 cm
22.5 cm
30 cm
SOLUTION
Solution : B
ABDE =ACDF = BCEF = 24 = 12
DE=2×AB=6 cm,DF=2×AC=5 cm.
∴ Perimeter of ΔDEF=(DE+EF+DF)
=6+4+5=15 cm.
Question 2
ABC is a triangle and DE is drawn parallel to BC cutting the other sides at D and E. If AB=3.6 cm,AC=2.4 cm and AD=2.1 cm, then AE = ______.
SOLUTION
Solution : A
By Basic Proportionality Theorem,
BDAD=CEAE
BDAD+1=CEAE+1
AD+BDAD=AE+CEAE
ABAD=ACAE
Subtituting values:
3.62.1=2.4AE
AE=2.1×2.43.6=1.4 cm
Question 3
In the adjoining figure, if BC=a,AC=b,AB=c and ∠CAB=120∘,
then which of the following is the correct relation?
a2 = b2 + c2 + 2bc
a2 = b2 + c2 - 2bc
a2 = b2 + c2 + bc
a2 = b2 + c2 - bc
SOLUTION
Solution : C
In △CDB,
BC2=CD2+BD2 [Pythagoras theorem]
BC2=CD2+(DA+AB)2
BC2=CD2+DA2+AB2+(2×DA×AB) ...(i)
In △ADC,
CD2+DA2=AC2 ...(ii) [Pythagoras Theorem]
Here, ∠CAB=120∘ (given)
⇒∠CAD=60∘ (since ∠CAD and ∠CAB form a linear pair of angles)Also, cos60∘=ADAC
AC=2AD ...(iii)
Substituting the values from (ii) & (iii) in (i) we get,
BC2=AC2+AB2+(AC×AB)
a2=b2+c2+bc
Alternatively,
Since ∠A is an obtuse angle in △ABC so,
BC2=AB2+AC2+2AB.AD =AB2+AC2+2×AB×12×AC [∵AD=ACcos60∘=12AC] =AB2+AC2+AB×AC
⇒a2=b2+c2+bc
Question 4
In △ ABC, ∠A and ∠C are given, the number of triangles that can be constructed with this data is -
One
Two
Three
Infinite
SOLUTION
Solution : D
∠S=180o−(∠A+∠C)
The angles of the traingle are known. With the given measurement of the angles, we can draw infinite number of traingles (Similar triangles).
Question 5
In triangle ABC, D is a point on AB and E is a point on AC such that DE || BC. If ADAB = AEx, Then x is
SOLUTION
Solution :In △ABC,
DE ∥ BC (Given)
Therefore, ADAB = AEAC [By Basic Proportionality Theorem]
Comparing the above with ADAB = AEx
x = AC
Question 6
In below shown figure, PSSQ= PTTR and∠PST= ∠PRQ. Then ΔPQR is a/an
SOLUTION
Solution :It is given that
PSSQ = PTTR
So, ST||QR [By converse of Basic Proportionality Theorem]
∴, ∠PST=∠PQR (Corresponding Angles)
Also, it is given that
∠PST = ∠PRQ
So,∠PRQ = ∠PQR
Therefore, PQ = PR (Sides opposite the equal angles)
i.e., ΔPQR is an isosceles triangle.
Question 7
ABCD is a square. Equilateral triangles ACF and ABE are drawn on the the diagonal AC and side AB respectively. Find area of △ACF : area of △ABE.
√2:1
2:1
4:1
8:1
SOLUTION
Solution : B
Let AB be a units long.
We know that diagonal of a square = √2× side length
⇒AC=√2×ABNow, △ ABE and △ ACF are equilateral triangles.
Each angle of both △ ABE and △ ACF is 60∘.
∴△ABE∼△ ACF (by AAA similarity criterion)
So, ar(ACF)ar(ABE)=AC2AB2=2a2a2=21
Question 8
Find the ratio of area of the equilateral triangles whose sides are 3 and 4 units.
9 : 25
9: 16
16: 25
SOLUTION
Solution : B
Let the triangles be △ABC and △DEF
Then AB=3 units and DE=4 units.Each angle of both △ABC and △DEF is 60∘.
△ABE∼△ ACF
( AAA similarity)So,
ar(ABC)ar(DEF)=BC2EF2=3242=916
Question 9
In a right △ABC, a perpendicular BD is drawn on to the largest side from the opposite vertex. Which of the following does not give the ratio of the areas of △ABD and △ACB?
(ABAC)2
(ADAB)2
(BDCB)2
(ABAD)2
SOLUTION
Solution : D
Consider ΔABD and ΔACB:
∠BAD = ∠BAC [common angle]
∠BDA = ∠ABC [ 90∘]
By AA similarity criterion,
△ ABD ~ △ACBHence,
ar(ΔABD)ar(ΔACB)=(ABAC)2=(ADAB)2=(BDCB)2
Question 10
A ladder is resting on a wall of height 10√7m such that the foot of the ladder when placed 10√7m away from the wall, half of the ladder is extending above the wall. When the tip of the ladder is placed on the tip of the wall, how far is the foot of the ladder from the wall?
100 √7 m
70√7 m
70 m
70√11 m
SOLUTION
Solution : C
Consider the above figure:
AD is the part of the ladder in the first case and AC is the ladder in the second case
AB = BD = 10 √7m [given]
2AD = AC [since AD is the half of the ladder AC]
∴ By Pythagoras Theorem in △ABD,
AD2 = AB2 + BD2
⇒ AD2 = (10√7)2 +(10√7)2
= 2 (10√7)2⇒AD=10√7√2
Since AD is half of the ladder, the complete length of the ladder
AC = 2AD = 20 √2√7
Now,
In ΔABC,
AC2 = AB2 + BC2⇒ BC2=AC2−AB2
=(20√7√2)2 - (10√7)2
=(10√7)2[(2√2)2−1]
= (10√7)2×7∴BC=10√7×√7=70 m