Exercise 6.4
Class X
NCERT Textbook
Q.No. 1. Let △ ABC ∼△DEF and there areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.
Solution:
ar(ABC)/ar(DEF) = (BC/EF)²
Now, putting the value of areas and EF
64/121 = (BC/15.4)²
Therefore,
BC = 8x15.4/11
Hence,
BC = 11.2 cm Ans.
Q.No. 2. Diagonals of trapezium ABCD with AB॥DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.
Solution:
Consider in triangles AOB and COD.
<OAB = <OCD (Alternate angles)
<OBA = <ODC (Alternate angles)
<AOB = <COD (Vertically opposite angles)
△AOB∼ △COD (AAA Similarity)
Therefore,
ar(AOB)/ar(COD) = (AB/CD)²
Now, putting the value of AB.
ar(AOB)/ar(COD) = (2CD/CD)²
ar(AOB)/ar(COD) = 4/1
Required ratio = 4:1 Ans.
Q.No.3. In Fig. 6.46 ABC and DBC are two triangles on the same base BC. If AD intersect BC at O, show that
ar(ABC)/ar(DBC) = AO/DO
Solution:
Draw perpendicular AP and DP on BC.
Consider in triangles APO and DQO.
<APO = <DQO (Each 90⁰)
<AOP = <DOQ (Vertically opposite angles)
△APO ∼ △DQO
Therefore,
ar(APO)/ar(DQO) = (AP/DQ)² = (AO/DO)² = (PO/QO)² -------------- (i)
Again,
ar(ABC) = BCxAP/2
ar(DBC) = BCxDQ/2
ar(ABC)/ar(DBC) = AP/DQ
Hence from (i),
ar(ABC) = ar(DBC) = AO/DO Proved.
Q.No. 4. If the areas of two similar triangles are equal, prove that they are congruent.
Solution:
Consider about two triangles ABC and PQR.
Therefore,
ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)² ------------------ (i)
Again,
ar(ABC) = ar(PQR) (Given) ---------------------- (ii)
Now, from (i) and (ii).
AB = PQ
BC = QR
AC = PR
Hence,
△ABC≅△PQR Proved.
Q.No. 5. D, E and F respectively the mid points of the sides AB, BC and CA of triangle ABC. Find the ratio of the areas of △ DEF and △ABC.
Solution:
D and F are the mid points of AB and CA.
Therefore,
AD/DB = AF/FC
So,
DF ॥ BC
and
DF = BC/2 ------------- (i)
Similarly,
DE = AC/2 ------------- (ii)
and
EF = AB/2 ----------- -- (iii)
Now, from (i), (ii) and (iii).
ADEF, BEFD and CFDE are parallelograms.
Now, consider in triangles ABC and DEF.
<A = <E ---------------- (ADEF is a parallelogram)
<B = <F ---------------- (BEFD is a parallelogram)
<C = <D ----------------- ( CFDE is parallelogram)
Hence,
△ABC ∼ △EFD
ar(ABC)/ar(DEF) = (AB/EF)²
Now, putting the value of EF.
ar(ABC)/ar(DEF) = 4:1
Therefore,
ar(DEF)/ar(ABC) = 1:4 Ans.
Q.No. 6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Solution:
Consider about two similar triangles ABC and PQR.
AX, BY and CZ are the medians of ABC. PK, QL and RM are medians of PQR.
Since,
ABC and PQR are similar triangles.
ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)² ---------------------- (i)
Since,
AX and PK are medians of ABC and PQR.
Therefore,
Triangles ABX and PQK are similar.
ar(ABX)/ar(PQK) = (AB/PQ)² = (AX/PK)² ---------------------- (ii)
Now, from (i) and (ii).
The ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding medians. Proved.
7. Prove that area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Solution:
Let the side of the square be X.
Diagonals of the square = ✔️2 X
Area of equilateral triangle on diagonal = 2✔️3X²/4 ------- (i)
Area of equilateral triangles on side = ✔️3X²/4 ------ (ii)
Area of equilateral triangles on side = ✔️3X²/4 ------ (ii)
Now, from (i) and (ii)
Area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Proved
