Kerala Syllabus Class 7 Mathematics - Unit 11 Squares and Right Triangles - Questions and Answers | Teaching Manual 

Questions and Answers for Class 7 Mathematics - Unit 11 Squares and Right Triangles - Study Notes | Text Books Solution STD 7 - Maths: Unit 11 Squares and Right Triangles - Questions and Answers | ഗണിതം - സമചതുരങ്ങളും മട്ടത്രികോണങ്ങളും - ചോദ്യോത്തരങ്ങൾ 

ഏഴാം ക്ലാസ്സ്‌  Mathematics - Unit 11 Squares and Right Triangles എന്ന പാഠം ആസ്പദമാക്കി തയ്യാറാക്കിയ ചോദ്യോത്തരങ്ങള്‍. ഈ അധ്യായത്തിന്റെ Teachers Handbook, Teaching Manual എന്നിവ ഡൗൺലോഡ് ചെയ്യാനുള്ള ലിങ്ക് ചോദ്യോത്തരങ്ങളുടെ അവസാനം നൽകിയിട്ടുണ്ട്. പുതിയ അപ്‌ഡേറ്റുകൾക്കായി ഞങ്ങളുടെ Telegram Channel ൽ ജോയിൻ ചെയ്യുക.

Kerala Syllabus STD 7 Maths: Unit 11 Squares and Right Triangles - Textbook Solutions

♦ Textbook Activities (Textbook Page No: 149)
♦ Now compute the lengths of the sides of squares with area given below. Write the answers using square roots.
(i) 49 square centimetres 
Answer:
Area of the square = 49 sq.cm
Length of the side = √49 = √7² = 7cm

(ii) 169 square centimetres 
Answer:
Area of the square = 169 sq.cm
169 = 13 x 13 = 13²
Length of the side = √169 = √13² = 13cm

(iii) 400 square centimetres 
Answer:
Area of the square = 400 sq.cm
400 = 2 x 2 x 2 x 2 x 5 x 5
       = 2⁴ x 5²
       = (2²)² x 5² = 4² x 5²
       = (4 x 5)² = 20²
Length of the side = √400 = √20² = 20cm

(iv) 225 square centimetres 
Answer:
Area of the square = 225 sq.cm
225 = 5 x 5 x 3 x 3
       = 5² x 3² = (5 x 3)² = 15²
Length of the side = √225 = √15² = 15cm

(v) 1.69 square centimetres 
Answer:
Area of the square = 1.69 sq.cm
1.69 = 169/100 = 13²/10² = (¹³⁄₁₀)² = (1.3)²
Length of the side = √1.69 = √(1.3)² = 1.3cm

(vi) 6¼ square metres
Answer:
Area of the square = 6¼ sq.cm
6¼ = ²⁵⁄₄ = 5²/2² = (⁵⁄₂)² 
Length of the side = √6¼ = √(⁵⁄₂)² = ⁵⁄₂ = 2½ m
♦ Textbook Activities (Textbook Page No: 152)
♦ Now, can’t you draw squares of areas as below:
(i) 32 sq.cm 
Answer:
32 ÷ 2 = 16
Side of the square having area 16 sq.cm is √16 = 4 cm
The length of one side of the small square is 4 cm
Then extend one side and mark the length of the diagonal on it, and construct the square with an area of 32 sq.cm.
(ii) 50 sq.cm 
Answer:
50 ÷ 2 = 25 
Side of the square having area 25 sq.cm is √25 = 5 cm
Draw a square having side 5 cm with area 25 sq.cm. Then extend one side and mark the length of the diagonal on it, and construct the square with an area of 50 sq.cm.
Length of one side of the small square is 5 cm
(iii) 12.5 sq.cm  
Answer:
12.5 ÷ 2 = 12.5/2 = ¹²⁵⁄₂₀ = 6.25
6.25 = ⁶²⁵⁄₁₀₀ = 25²/10² = (²⁵⁄₁₀)² = (2.5)²
The side of a square having area 6.25 sq.cm is 2.5 cm
Draw a square having side 2.5 cm and then extend one side and mark the length of the diagonal on it, and construct the square with an area 12.25 sq.cm.
(iv) 24½ sq.cm
Answer:
24½ ÷ ⁴⁹⁄₂, ⁴⁹⁄₂ + 2 = ⁴⁹⁄₂ x ½ = ⁴⁹⁄₄ = 7²/2² = (⁷⁄₂)²
√(⁷⁄₂)² = ⁷⁄₂ = 3½ = 3.5 cm
The side of a square having area ⁴⁹⁄₄ sq.cm is ⁷⁄₂ cm
Draw a square having side 3.5 cm and then extend one side and mark the length of the diagonal on it, and construct the square with an area 24½ sq.cm.
♦ Textbook Activities (Textbook Page No: 157, 158)
♦ Now try these problems:
(1) Draw squares of areas given below
(i) 17 square centimetres
Answer:
17 = 1 + 16 = 1² + 4²
By using Pythagoras Theorem, we can draw a right triangle with perpendicular sides 4cm and 1cm. Then the square by taking the hypotenuse as side gives the required square.
(ii) 18 square centimetres
Answer:
18 = 9 + 9 = 3² + 3²
By Pythagoras Theorem, draw a right triangle with perpendicular sides 3cm and 3cm. Then, draw a square by taking the hypotenuse as a side, we get the square of area 18 sq cm.
(iii) 19 square centimetres
Answer:
19 = 100 - 81 = 10² - 9²
By Pythagoras Theorem, if we draw a right triangle with a hypotenuse of 10 cm and another side of 9 cm. Then the area of the square drawn on the third side would be 19 sq cm.
(2) In the picture below, dots are marked 1 centimetre apart, horizontally and vertically and some of these are joined to make squares:
Calculate their areas.
Answer:
• Area of fig. (1) = 3² + 2² = 9 + 4 = 13 sq cm
• Area of fig. (2) = 1² + 1² = 1 + 1 = 2 sq cm
• Area of fig. (3) = 2² + 1² = 4 + 1 = 5 sq cm
• Area of fig. (4) = 4² + 2² = 16 + 4 = 20 sq cm
♦ Textbook Activities (Textbook Page No: 159, 160)
♦ Now try these problems:
(1) The lengths of two sides of some right triangles are given below. Calculate the length of the third side.
(i) Perpendicular sides 6 centimetres, 8 centimetres
Answer:
By adding the squares of the perpendicular sides, we get the square of the hypotenuse.
Square of the hypotenuse = 6² + 8² = 36 + 64 = 100
∴ Length of the hypotenuse = √100 = 10 cm
(ii) Perpendicular sides 9 centimetres, 12 centimetres
Answer:
Square of the hypotenuse = 9² + 12² = 81 + 144 = 225
∴ Length of the hypotenuse = √225 = 15 cm
(iii) Perpendicular sides 7 centimetres, 24 centimetres
Answer:
Square of the hypotenuse = 7² + 24² = 49 + 576 = 625
∴ Length of the hypotenuse = √625 = 25 cm
(iv) Perpendicular sides 14 centimetres, 48 centimetres
Answer:
Square of the hypotenuse = 14² + 48² = 196 + 2304 = 2500
∴ Length of the hypotenuse = √2500 = 50 cm
(v) Hypotenuse 17 centimetres, another side 15 centimetres
Answer:
Square of the third side = 17² - 15² = 289 - 225 = 64
∴ Length of the third side = √64 = 8 cm
(vi) Hypotenuse 34 centimetres, another side 30 centimetres
Answer:
Square of the third side = 34² - 30² = 1156 - 900 = 256
∴ Length of the third side = √256 = 16 cm
(2) Two pillars of heights 2 metres and 5 metres stand 4 metres apart:
What is the distance between their tops?
Draw a perpendicular to BD from E
AB = CE = 4 m; BC = AE = 2 m; BD = 5 m
CD = BD - BC = 5 - 2 = 3 m
In the fig. triangle DCE is a right triangle.
ED² = CE² + CD² = 4² + 3² = 16 + 9 = 25
Square of the hypotenuse = 25
Length of the hypotenuse, ED = √25 = 5 m
∴ Distance between the tops of the pillars = 5 m

(3) Find the length of the bottom side of the triangle drawn below:
In the figure, AB = AC = 13 cm
AD = 12 cm
Here, we divide ∆ABC into two right-angled triangles.
In right triangle ABD,
BD² = 13² - 12² = 169 - 144 = 25
BD = √25 = 5 cm
In right triangle ACD,
CD² = 13² - 12² = 169 - 144 = 25
CD = √25 = 5 cm
BC = BD + CD = 5 + 5 = 10 cm
Bottom side of the triangle = 10 cm

(4) In the pictures below, O is the centre of the circle and A, B, P, Q are points on the circle:
Calculate the lengths of the lines AB and PQ.
In the figure, OB = OA = 5 cm (radii)
OP = 4 cm
In right triangle APO, AP² = 5² - 4² = 25 - 16 = 9
AP = √9 = 3 cm; BP = 3 cm
AB = AP + BP = 3 + 3 = 6 cm
Length of AB = 6 cm
In the figure, OP = OQ = 5 cm (radii)
OP = 3 cm
In right triangle ORP, PR² = 5² - 3² = 25 - 9 = 16
PR = √16 = 4 cm; QR = 4 cm
PQ = PR + QR = 4 + 4 = 8 cm
Length of PQ = 8 cm