Mathematics Part ii Solutions Solutions for Class 7 Math Chapter 2 Triangular Area are provided here with simple step-by-step explanations. These solutions for Triangular Area are extremely popular among class 7 students for Math Triangular Area Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part ii Solutions Book of class 7 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part ii Solutions Solutions. All Mathematics Part ii Solutions Solutions for class 7 Math are prepared by experts and are 100% accurate.
Page No 108:
Question 1:
Now can’t you find the areas of these triangles?
Answer:
(i) Let the given triangle be ΔABC.
We know that the area of a right-angled triangle is half the product of its perpendicular sides.
∴ Area of ΔABC =
(ii) Let the given triangle be ΔABC.
We know that the area of a right-angled triangle is half the product of its perpendicular sides.
∴ Area of ΔABC =
(iii) Let the given triangle be ΔABC.
We know that the area of a right-angled triangle is half the product of its perpendicular sides.
∴ Area of ΔABC =
(iv) Let the given triangle be ΔABC.
We know that the area of a right-angled triangle is half the product of its perpendicular sides.
∴ Area of ΔABC =
Page No 109:
Question 1:
Find the areas of these figures also:
Answer:
(i) The named figure is shown below.
We know that the area of a rectangle is the product of the lengths of its adjacent sides.
Area of the rectangle AECF = FC × EC = 4 cm × 3 cm = 12 cm2
We know that the area of a right-angled triangle is half the product of its perpendicular sides.
∴ Area of ΔADF =
Area of ΔCEB =
Area of the given figure = Area of ΔADF + area of rectangle AECF + area of ΔCEB
= (3 + 12 + 3) cm2
= 18 cm2
(ii) The named figure is shown below.
We know that the area of a rectangle is the product of the lengths of its adjacent sides.
Area of rectangle ABFE = EF × AE = 5 cm × 2 cm = 10 cm2
We know that the area of a right-angled triangle is half the product of its perpendicular sides.
∴ Area of ΔADE =
Area of ΔBFC =
Area of the given figure = Area of ΔADE + area of rectangle ABFE + area of ΔBFC
= (1 + 10 + 1) cm2
= 12 cm2
(iii) The named figure is shown below.
We know that the area of a right-angled triangle is half the product of its perpendicular sides.
∴ Area of ΔADB =
Area of ΔADC =
Area of the given figure = Area of ΔADB + area of ΔADC
= (3 + 3) cm2
= 6 cm2
(iv) The named figure is shown below.
We know that the area of a right-angled triangle is half the product of its perpendicular sides.
Area of ΔADB =
Area of ΔADC =
Area of the given figure = Area of ΔADB + area of ΔADC
= (4 + 6) cm2
= 10 cm2
Page No 113:
Question 1:
Now can’t you find the areas of these triangles?
Answer:
(i) Let the given triangle be ΔABC and AD be the altitude of ΔABC.
We know that the area of a triangle is half the product of any side and the altitude to it.
∴ Area of ΔABC =
(ii) Let the given triangle be ΔABC and CD be the altitude of ΔABC.
We know that the area of a triangle is half the product of any side and the altitude to it.
∴ Area of ΔABC =
(iii) Let the given triangle be ΔABC and AD be the altitude of ΔABC.
We know that the area of a triangle is half the product of any side and the altitude to it.
∴ Area of ΔABC =
Page No 113:
Question 2:
Compute the altitudes to the other two sides of this triangle:
Answer:
Let the given triangle be ΔABC and AD, BE, CF be the three altitudes of ΔABC.
We know that the area of a triangle is half the product of any side and the altitude to it.
∴ Area of ΔABC =
Also, area of ΔABC =
⇒ 6 cm2 = cm
⇒ BE = = 2.4 cm
Again, area of ΔABC =
⇒ 6 cm2 = cm
⇒ CF = = 3 cm
Thus, the lengths of the other altitudes are 2.4 cm and 3 cm.
View NCERT Solutions for all chapters of Class 7