10th Maths 6.4
NCERT Class 10th solution of Exercise 6.1
NCERT Class 10th solution of Exercise 6.2
NCERT Class 10th solution of Exercise 6.3
NCERT Class 10th solution of Exercise 6.5
Exercise 6.4
Q1. Let △ABC∼△DEF and their areas be, respectively, 64cm2 and 121cm2. If EF=15.4 cm, find BC.
Sol.
Given,
△ABC∼△DEF
so that
ABDE=BCEF=CAFD
ar(ABC)ar(DEF)=(BCEF)2
64121=(BC15.4)2
(811)2=(BC15.4)2
BC=8×15.411
BC=123.211
BC=11.2 cm
Answer:
BC=11.2 cm.
Q2. Diagonals of a 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.
Sol.
Given,
ABCD is trapezium with AB||DC and AB = 2CD
In△AOB and△COD
∠1=∠2 [Alternate angles]
∠AOB = ∠COD [Verticaly opposite angles]
△AOB∼△COD [By AA]
So that
ar(AOB)ar(COD)=(ABCD)2
=(2CDCD)2
=41
=4:1
Anser:
ar(AOB) : ar(COD) = 4:1
Q3. In Figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, Show that ar(ABC)ar(DBC)=AODO.
Sol.
Given,
△ABC and △DBC on the same base BC and AD intersects BC at O
ar(ABC)ar(DBC)=12×BC×AL12×BC×DM
AODO=ALDM______(1)
In△ALO and △DMO
∠ALO=∠DMO=90∘[AL⊥BC,DM⊥BC]
∠AOL=∠DOM[Vertically opposite angles]
△ALO∼△DMO [By AA]
So that
AODO=ALDM[from (1)]
Proved
Q4. If the area of two similar triangles are equal, prove that they are congruent.
Sol.
Given
△ABC∼△PQR
and
ar(ABC)=ar(PQR)
To Prove:△ABC≅△PQR
Proof:
ar(ABC)ar(PQR)=(ABPQ)2=(BCQR)2=(CARP)2
ar(ABC)ar(ABC)=(ABPQ)2=(BCQR)2=(CARP)2
1=
(text{AB})=(text{PQ})
(text{BC})=(text{QR})
(text{CA})=(text{RP})
triangle text{ABC}cong triangle text{PQR}text{ [By SSS}]
text{Proved}
Q5. D, E and F are respectively the mid-points of sides AB, BC and CA of triangle ABC. Find the ratio of the area of triangle DEF and triangle ABC.
text{Sol.}
text{In }triangle text{ABC},text{D, E, and F are mid-points }text{of sides AB, BC and CA respectively}
text{DF}=1/2text{BC}text{ [By Midpoint Theorem]}
text{DE}=1/2text{AC}text{ [By Midpoint Theorem]}
text{EF}=1/2text{AB}text{ [By Midpoint Theorem]}
(text{DF})/(text{BC})=(text{DE})/(text{AC})=(text{EF})/(text{AB})=1/2
text{Therefore}
triangle text{ABC}∼triangle text{DEF}text{ [By SSS]}
text{And}
(ar(text{ABC}))/(ar(text{DEF}))=((text{DF})/(text{BC}))^2
(ar(text{ABC}))/(ar(text{DEF}))=(1/2)^2
ar(text{ABC}):ar(text{DEF})=1:4
text{Answer:}
ar(text{ABC}):ar(text{DEF})=1:4
Q6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
text{Sol.}
text{Given}
triangle text{ABC}∼triangle text{PQR}
text{AL and PM are medians}
text{To Prove}
(ar(text{ABC}))/(ar (text{PQR})) = ((text{AL})/(text{PM}))^2
text{Proof}
text{In}triangle text{ABC and }triangle text{PQR}
(ar (text{ABC}))/(ar (text{PQR})) = ((text{AB})/(text{PQ}))^2
triangle text{ABC}∼triangle text{PQR}
(text{AB})/(text{PQ})=(text{BC})/(text{QR})=(text{2BL})/(text{2QM})
(text{AB})/(text{PQ})=(text{BL})/(text{QM})____(1)
angle text{B}=angle text{Q}[text{angle oppsite to equal side}]
triangle text{ABL}∼triangle text{PQM}
(text{BL})/(text{QM})=(text{AL})/(text{PM})____(2)
text{from equation (1) and (2)}
(text{AB})/(text{PQ})=(text{AL})/(text{PM})
(ar (text{ABC}))/(ar (text{PQR})) = ((text{AL})/(text{PM}))^2
text{Proved}
Q7. Prove that the 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.
text{Sol.}`
text{Let PQRS is a square of side length a.}
text{and diagonal PR }asqrt2
triangle text{QAR}=triangle text{PRT are equilateral triangles, }text{so these are similar}
(ar(text{QAR}))/(ar(text{PRT}))=((text{QR})/(text{PR}))^2
=a^2/(asqrt2)^2
=a^2/(2a^2)
=1/2
ar(text{QAR})=1/2ar(text{PRT})
text{Proved}.
Tick the correct answer and justify:
Q8.ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is
A) 2:1
B) 1:2
C) 4:1
D) 1:4
text{Answer:}
C) 4:1
text{Justifiction}
text{Let Sides of equilateral triangles }text{AB=BC=CA=a}
text{D is mid point of BC}
text{DB}=1/2a
triangle(text{ABC})∼triangle(text{BDE})
(ar(text{ABC}))/(ar(text{BDE}))=((text{BC})/(text{BD}))^2
=a^2/(1/2a)^2
=a^2/(1/4a^2)
=4/1
=4:1
Q9. Sides of two similar triangles are in the ratio 4:9. Areas of these triangles are in the ratio
A) 2:3
B) 4:9
C) 81:16
D) 16:81
text{Answer:}
D) 16:81
text{Justification}
text{Ratio of area of triangles = }(4/9)^2=16/81
Comments