1) Find the perimeter of the rectangle whose length is 40cm and a diagonal is 41 cm.
2) The diagonals of a rhombus measure 16cm and 30cm. Find its perimeter.
Answers
Step-by-step explanation:
perimeter perimeter of the rectangles 2 x length + breadth
2 x 40 cm + 41 cm
3280cm answer
2. diagonal of a rhombus = 1/2x product of diagonals
1/2x480cm
240 cm answer
Answer:
ans 1) : 98 cm
ans 2) : 68 cm
Step-by-step explanation:
ans 1) : Let the triangle be ABCD
Where,
length = AB = 40 cm
breadth = BC
diagonal AC = 41 cm
perimeter of ABCD = 2 (length +breadth)
To find breadth BC,
in rectangle,all angles are 90°,
angle A = angle B = angle C = angle D = 90°
therefore,
in right triangle ABC,
By phythagorus theorem,
AC square = BC square + AB square
(41 cm) square= BC square + (40 cm) square
1681 cm = BC square + 1600 cm
1681 cm - 1600 cm = BC square
81 cm = BC square
( 9) square = BC square
Cancelling squares :
= BC = 9 cm
therefore breadth = 9 cm
now,
perimeter of a rectangle = 2 ( length + breath)
= 2 ( 40 cm + 9 cm)
= 2 × 49cm
= 98 cm...
ans 2) : let ABCD be the given rhombus,
where,
BD = 16 cm
AC = 30 cm
we know that,
Diagonals of a rhombus are perpendicular to each other
therefore AC is perpendicular to BD
and OB = BD/2 = 16/2 = 8 cm
OA = AC/2 = 30/2 = 15 cm
now,
in triangle AOB, right angled at O
By Pythagoras theorem,
AB square = OA square + OB square
AB square = (15) square + (8) square
AB square = 225 + 64
AB square = 289
AB square = (17)square
Cancelling squares :
AB = 17 cm
now,
perimeter ABCD = AB + BC + CD + AD
= AB + AB + AB + AB ( as value of all sides are equal)
= 4 AB
= 4 × 17
= 68 cm....