Question 1 ——————————-
Answer 1:
Given, Angle EOB = 90, AOD = 40
From the diagram, it is clear that AOB is a straight(St.) line.
Angle of st. line = 180 deg
or, AOD + DOE + EOB = 180
or, 40 + b + 90 = 180
or, b = 50 deg. is the answer

Answer 2.1:
Sum of angles in a triangle = 180
For angle PQR, angle P + angle Q + angle R = 180

From the diagram, angle P = w,
angle Q = w+20
angle R = 2w+20

2. To classify triangle TQR
Given, QT = TR = 13m (We calculated in 1.)
The third side of the triangle TQR is QR.
Length of QR = Length of PT + length of TS
or, QR = 5+9 = 14m.

Therefore, in triangle TQR, two sides are equal, QT = TR. So it is an isosceles triangle.

3. To find the area of triangle TSR

Area of triangle = 1/2 * (base * height)
or, Area = 1/2* (TS * SR)
or, Area = 1/2* 9 * 12
or, Area = 54 m^2

Angle subtended by an arc Source Rate the problemSpam (0)Hard (0)Helpful (0)Wow (0) :57:31+00:00kishorGeometrySourceKishor Kumarhi.friend_zone@rocketmail.comSubscriberSOLVE - ΜΔΓΗŠ

Sides of a square Source Rate the problemHelpful (1)Spam (0)Hard (0)Wow (0) :48:51+00:00kishorGeometry Source Kishor #101;@rocketmail.comSubscriberSOLVE - ΜΔΓΗŠ

calculate angle X The two straight lines CD and AB are parallel to each OAB = 60 OCD = 110 Angle X. Rate the problemSpam (0)Hard (0)Helpful (0)Wow (0) :31:36+00:00kishorGeometryThe two straight...

Maximum value of 2x+3y If (x,y) lies on a circle of a radius 5 with the origin as its center, find the maximum value of (2x + 3y) Rate the problemHelpful (1)Spam (0)Hard (0)Wow (0) :03:32+00:00RajniCalculusGeometryIf (x,y)...

How many coins can cover the entire table? A rectangular table has 100 coins placed on it (centers must be on the table) such that none of the coins overlap, and it is impossible to place any...

Question 1 ——————————-

Answer 1:

Given, Angle EOB = 90, AOD = 40

From the diagram, it is clear that AOB is a straight(St.) line.

Angle of st. line = 180 deg

or, AOD + DOE + EOB = 180

or, 40 + b + 90 = 180

or, b = 50 deg. is the answer

Answer 2.1:

Sum of angles in a triangle = 180

For angle PQR, angle P + angle Q + angle R = 180

From the diagram, angle P = w,

angle Q = w+20

angle R = 2w+20

Therefore, w + (w+20) + (2w+20) = 180

or, 4w + 40 = 180

or, 4w = 140

or, w = 35 deg.

Answer 2.2:

Angle R = 2w+20

We calculated w = 35 deg.

Thus, angle R = 2*35 + 20

angle R = 90 deg.

Question 2 ——————————-

1. To find the value of C

Pythagoras Theorem: (Hypotenuse)^2 = (perpendicular)^2 + (base)^2

Following this,

(TQ)^2 = (PQ)^2 + (PT)^2

or, c^2 = 12^2 + 5^2

or, c^2 = 144 + 25

or, c^2 = 169

or, c = sqrt(169)

or, c = 13m

2. To classify triangle TQR

Given, QT = TR = 13m (We calculated in 1.)

The third side of the triangle TQR is QR.

Length of QR = Length of PT + length of TS

or, QR = 5+9 = 14m.

Therefore, in triangle TQR, two sides are equal, QT = TR. So it is an isosceles triangle.

3. To find the area of triangle TSR

Area of triangle = 1/2 * (base * height)

or, Area = 1/2* (TS * SR)

or, Area = 1/2* 9 * 12

or, Area = 54 m^2