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Find the measure of the angle x in the given figure.
∠EFD+∠FED = x
(Exterior angle property of a triangle)
⇒ 28o+42o = ∠x or ∠x = 70o
Which of the following options is INCORRECT?
∠1 =∠2 and ∠2 =∠3 [Alternate angles]
So, ∠1 =∠3 and ∠1+∠4+∠5 = 180o [Angle sum property]
Also, ∠8 = ∠6
In the figure (not drawn to scale), ADF and BEF are triangles and EC = ED, find y.
In ΔCED, CE=ED
∴ ∠EDC = ∠ECD
[Angles opposite to equal sides are equal] ⇒ ∠ECD = 28o
Also, ∠ECD = ∠BCA (Vertically opposite angles) ⇒ ∠BCA = 28o
In ΔBCA, y = 62o+28o
Exterior angle property] ⇒ y = 90o
In a ΔABC,which of the given condition holds?
In the figure (not drawn to scale), ABC is an equilateral triangle and ABD is an isosceles triangle with DA = DB, find x.
Since ABC is an equilateral triangle.
∴ ∠CAB =∠ABC =∠BCA = 60o
And ∠DBA = ∠DAB = (60o−x)
[∵ DA = DB]
In ΔDAB, ∠DAB∠DAB+∠ADB =180o
⇒ 2(60o−x)+88o = 180o
⇒ 2(60o−x) = 92o
⇒ 60o−x = 46o ⇒ x = 14o
ABC is an isosceles triangle with AB = AC and AD is altitude, then ____.
In the figure (not drawn to scale), ABCD is a square, ADE is an equilateral triangle and BFE is a straight line, find y.
In ΔAEB,
∠A=∠DAE+∠BAD ⇒ ∠A=60o+90o=15o
And, AE=AB ⇒ ∠ABE=∠AEB
[Angles opposite to equal sides are equal]
Now, ∠A+∠ABE+∠AEB=180o (Angle sum property)
⇒ 2∠AEB=180o −150o = 30o ⇒ ∠AEB = 15o
Now, ∠E=60o ⇒ ∠DEF=60o−15o = 45o
∴ In ΔEFD, ∠DEF+∠EDF+∠EFD
= 180o ⇒ 45o+60o+y = 180o
⇒ y = 180o−(45o+60o) = 75o
Find the measure of the angle x in the given figure.
∠UXV = y (Vertically opposite angles)
∴ y = 45o In ΔXYZ y+x+63o = 180o
(Angle sum property)
⇒ 45o+x+63o = 180o
⇒ x = 180o−(45o+63o)
⇒ x= 180o−108o = 72o
The given figure shows three identical squares. Find x.
We have, ABCD, CEFG and CIHJ are all squares.
So, ∠1+∠2+x=90o ???
(i) 36o+∠1+x=90o q ??.
(ii) x+∠2+27o=90o ??.
(iii) Adding (ii) and (iii),
we get 36o+x+27o+(∠A+∠2+x)=180o
⇒ 63o+x+90o=180o
(From (i)) ⇒ x=180o−153o = 27o
In the figure (not drawn to scale), EFA is a right-angled triangle with ∠EFA=90o and FGB is an equilateral triangle, find y−2x.
In ΔFGC, ∠CBF = 60o
(Angle of equilateral triangle)
∴ x+60o+92o = 180o
⇒ x = 180o−152o= 28o
Now, In ΔBCF, ∠CBF = 60o
∠FCB = 180o−92o (Linear pair)
⇒ ∠FCB=88o
∴ ∠BFC+88o+60o=180o
(Angle sum property)
⇒ ∠BFC=180o−148o = 32o
And ∠AFE = 90o ⇒ y+32o = 90o
⇒ y = 90o−32o = 58o
∴ y−2x = 58o−2×28o = 58o−56o = 2o
In the figure (not drawn to scale), ABC and DEF are two triangles, CA is parallel to FD and CFBE is a straight line. Find the value of x+y
∠FCA = ∠BFD
(Corresponding angles)
⇒ x = 51o
Now, in ΔABC y = 51o+83o (Exterior angle property)
⇒ y = 134o So, x+y = 51o+134o = 185o
In a ΔABC,if AB+BC = 10 cm, BC+CA = 12 cm,CA+AB = 16 cm, then the perimeter of the triangle is ____.
It is given that,
AB+BC = 10cm ?..(i)
BC+CA = 12cm ?..(ii)
CA+AB = 16cm ?..(iii)
Adding (i), (ii) and (iii); we get
2(AB+BC+CA) = 10+12+16
⇒ AB+BC+CA = 19cm
In the figure, not drawn to scale, ACDF is a rectangle and BDE is a triangle. Find ∠BED
∠CDB+∠BDE = 90o
(Angle of a rectangle)
⇒ 48o+∠BDE = 90o
⇒ ∠BDE = 90o−48o = 42o
In ΔBED ∠EBD+∠BDE+∠BED = 180o
(Angle sum property)
⇒ 86o+42o+∠BED = 180o
⇒ ∠BED = 180o−(86o+42o) = 52o
In the figure, ABCD is a square, MDC is an equilateral triangle. Find the value of x.
∆MDC is an equilateral triangle. therefore all the angles will be 60°.
angle MCB=BCD-MCD
=90-60 = 30°
NOW,
In ∆BCD,
BC=CD
Therefore angle XBC = 45°.
when angle sum is applied in ∆CBX, X=105°
Find the measure of ∠LNM in the given figure.
∠KLO = ∠MLN
∴ ∠MLN = 70o in ∠LMN,∠MLN+∠LNM+∠LMN = 180o
(Angle sum property)
⇒ 70o+∠LNM+50o=180o
⇒ ∠LNM = 180o−(70o+50o) = 60o
A 26 m long ladder reached a window 24 m from the ground on placing it against a wall. Find the distance of the foot of the ladder from the wall.
In ΔPRQ, PR2 = PQ2+QR2
(By Pythagoras theorem)
(26)2 =(24)2+QR2 or QR2
= 676−576 = 100
⇒ QR = √100 ⇒ QR = 10
∴ The distance of the foot of the ladder from the wall is 10 m.
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
Let KB is original height of the tree. In ΔABC,
AC2 = AB2+BC2 = 52+122 =25+144=169
∴ AC = √169 = 13m
KB = KA+AB = (13+5)m = 18m
∴ Original height of the tree is 18m.
Aryan wants to plant a flower on the ground in the form of a rhombus. The diagonals of the rhombus measures 42 cm and 56 cm. Find the perimeter of the field.
Since diagonals of a rhombus bisect each other at 90o .
Given: BD = 42 cm and AC = 56 cm
∴ BK = 1/2 BD = 42/2 = 21cm
AK=1/2 AC = 56/2 = 28 cm
In ΔKAB,AB2 = AK2+BK2 = (28)2+(21)2=784+441=1225
∴ AB=√1225 = 35cm
∴ Perimeter of the field ABCD = 4×35 = 140 cm;
A 34 m long ladder reached a window 16 m from the ground on placing it against a wall. Find the distance of the foot of the ladder from the wall.
Let AB = length of ladder, AC = height of window
In ΔABC, (AB)2 = (AC)2+(BC)2
⇒ (34)2 =(16)2+BC2
or
BC2 = (34)2−(16)2 ⇒ BC2 = 1156−256 = 900
∴ BC = √900 = 30m
Mrs Kaushik gives a problem to her students. Find the perimeter of a rectangle whose length is 28 cm and diagonal is 35 cm. What will be the correct answer?
ABCD is a rectangle,
In ΔACB, AC2 = AB2+BC2 (By Pythagoras theorem)
(35)2 = (28)2+BC2 or BC2 = (35)2−(28)2
⇒ BC2 = 1225−784 ⇒ BC2 = 441
∴ BC==21cm
∴ Perimeter of rectangle
= 2×(28+21)cm = 2×(49)cm = 98cm
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