10th Grade > Mathematics
TRIANGLES MCQs
Total Questions : 58
| Page 1 of 6 pages
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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.
:
In △ABC,
DE ∥ BC (Given)
Therefore, ADAB = AEAC [By Basic Proportionality Theorem]
Comparing the above withADAB = AEx
x = AC
Answer: Option A. -> 1:3
:
A
If two triangles are similar, then the ratio of the corresponding sides are equal.
Hence,
the ratio of heights = the ratio of sides = 1 : 3.
:
A
If two triangles are similar, then the ratio of the corresponding sides are equal.
Hence,
the ratio of heights = the ratio of sides = 1 : 3.
Answer: Option A. -> True
:
A
Both the figures are similar as all the corresponding angles are equal and all the corresponding sides are in the same ratio as all the sides of both the figures are equal.
Ratio of corresponding sides = 4.22.1=2
:
A
Both the figures are similar as all the corresponding angles are equal and all the corresponding sides are in the same ratio as all the sides of both the figures are equal.
Ratio of corresponding sides = 4.22.1=2
Answer: Option A. -> True
:
A
Pythagoras theorem states that a2+b2=c2
So he did correct i.e. he divided the land equally among his two sons.
:
A
Pythagoras theorem states that a2+b2=c2
So he did correct i.e. he divided the land equally among his two sons.
Answer: Option D. -> Infinite
:
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).
:
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).
Answer: Option B. -> 15 cm
:
B
ABDE =ACDF =BCEF =24 =12
DE=2×AB=6cm,DF=2×AC=5cm.
∴ Perimeter ofΔDEF=(DE+EF+DF)
=6+4+5=15cm.
:
B
ABDE =ACDF =BCEF =24 =12
DE=2×AB=6cm,DF=2×AC=5cm.
∴ Perimeter ofΔDEF=(DE+EF+DF)
=6+4+5=15cm.
Answer: Option C. -> 70 m
:
C
Consider the abovefigure:
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=70m
:
C
Consider the abovefigure:
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=70m